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Chemistry LibreTexts

5.3: Stoichiometry Calculations

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Learning Objectives

  • To balance equations that describe reactions in solution.
  • To calculate the quantities of compounds produced or consumed in a chemical reaction.
  • To solve quantitative problems involving the stoichiometry of reactions in solution.

A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. Stoichiometry is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. A stoichiometric quantity is the amount of product or reactant specified by the coefficients in a balanced chemical equation. This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?

All these questions can be answered using the concepts of the mole, molar and formula masses, and solution concentrations, along with the coefficients in the appropriate balanced chemical equation.

Stoichiometry Problems

When carrying out a reaction in either an industrial setting or a laboratory, it is easier to work with masses of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in and described in the following text.

Steps in Converting between Masses of Reactant and Product

  • Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass.
  • From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients).
  • Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are present in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess.

Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons).

To illustrate this procedure, consider the combustion of glucose. Glucose reacts with oxygen to produce carbon dioxide and water:

\[ C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{3.6.1} \]

Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain does not run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?

The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined above can be adapted as follows:

1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:

\[ moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose \nonumber \]

2. According to the balanced chemical equation, 6 mol of CO 2 is produced per mole of glucose; the mole ratio of CO 2 to glucose is therefore 6:1. The number of moles of CO 2 produced is thus

\[ moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber \]

\[ = 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber \]

\[ = 1.51 \, mol \, CO_2 \nonumber \]

3. Use the molar mass of CO 2 (44.010 g/mol) to calculate the mass of CO 2 corresponding to 1.51 mol of CO 2 :

\[ mass\, of\, CO_2 = 1.51 \, mol \, CO_2 \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.5 \, g \, CO_2 \nonumber \]

These operations can be summarized as follows:

\[ 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2 \nonumber \]

Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (Remember that you should generally carry extra significant digits through a multistep calculation to the end to avoid this!) This amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. Similar methods can be used to calculate the amount of oxygen consumed or the amount of water produced.

The balanced chemical equation was used to calculate the mass of product that is formed from a certain amount of reactant. It can also be used to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example \(\PageIndex{1}\), the mass of one reactant that is required to consume a given mass of another reactant.

Example \(\PageIndex{1}\): The US Space Shuttle

The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).

Image of a space shuttle taking off.

The US space shuttle Discovery during liftoff . The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine .

Given : reactants, products, and mass of one reactant

Asked for : mass of other reactant

  • Write the balanced chemical equation for the reaction.
  • Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons.

We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons.

A We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows:

\[ H_2 (g) + O_2 (g) \rightarrow H_2O (g) \nonumber \]

This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H 2 O and H 2 gives the balanced chemical equation:

\[ 2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g) \nonumber \]

Thus 2 mol of H 2 react with 1 mol of O 2 to produce 2 mol of H 2 O.

1. B To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:

\[ mass \, of \, O_2 = 1.00 \, tn \times { 2000 \, lb \over tn} \times {453.6 \, g \over lb} = 9.07 \times 10^5 \, g \, O_2 \nonumber \]

Using the molar mass of O 2 (32.00 g/mol, to four significant figures), we can calculate the number of moles of O 2 contained in this mass of O 2 :

\[ mol \, O_2 = 9.07 \times 10^5 \, g \, O_2 \times {1 \, mol \, O_2 \over 32.00 \, g \, O_2} = 2.83 \times 10^4 \, mol \, O_2 \nonumber \]

2. Now use the coefficients in the balanced chemical equation to obtain the number of moles of H 2 needed to react with this number of moles of O 2 :

\[ mol \, H_2 = mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} \nonumber \]

\[ = 2.83 \times 10^4 \, mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} = 5.66 \times 10^4 \, mol \, H_2 \nonumber \]

3. The molar mass of H 2 (2.016 g/mol) allows us to calculate the corresponding mass of H 2 :

\[mass \, of \, H_2 = 5.66 \times 10^4 \, mol \, H_2 \times {2.016 \, g \, H_2 \over mol \, H_2} = 1.14 \times 10^5 \, g \, H_2 \nonumber \]

Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors:

\[ tons \, H_2 = 1.14 \times 10^5 \, g \, H_2 \times {1 \, lb \over 453.6 \, g} \times {1 \, tn \over 2000 \, lb} = 0.126 \, tn \, H_2 \nonumber \]

The space shuttle had to be designed to carry 0.126 tn of H 2 for each 1.00 tn of O 2 . Even though 2 mol of H 2 are needed to react with each mole of O 2 , the molar mass of H 2 is so much smaller than that of O 2 that only a relatively small mass of H 2 is needed compared to the mass of O 2 .

Exercise \(\PageIndex{1}\): Roasting Cinnabar

Cinnabar, (or Cinnabarite) \(HgS\) is the common ore of mercury. Because of its mercury content, cinnabar can be toxic to human beings; however, because of its red color, it has also been used since ancient times as a pigment.

240px-Cinnabarit_01.jpg

Alchemists produced elemental mercury by roasting cinnabar ore in air:

\[ HgS (s) + O_2 (g) \rightarrow Hg (l) + SO_2 (g) \nonumber \]

The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction?

Calculating Moles from Volume

Quantitative calculations involving reactions in solution are carried out with masses , however, volumes of solutions of known concentration are used to determine the number of moles of reactants. Whether dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation give the number of moles of each reactant needed and the number of moles of each product that can be produced. An expanded version of the flowchart for stoichiometric calculations is shown in Figure \(\PageIndex{2}\). The balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used to determine the amounts of other species, as illustrated in the following examples.

The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.

Example \(\PageIndex{2}\) : Extraction of Gold

Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2 ] − ion. Gold is then recovered by reduction with metallic zinc according to the following equation:

\[ Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \nonumber \]

What mass of gold can be recovered from 400.0 L of a 3.30 × 10 −4 M solution of [Au(CN) 2 ] − ?

Given: chemical equation and molarity and volume of reactant

Asked for: mass of product

  • Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN) 2 ] − present by multiplying the volume of the solution by its concentration.
  • From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.

A The equation is balanced as written; proceed to the stoichiometric calculation. Figure \(\PageIndex{2}\) is adapted for this particular problem as follows:

As indicated in the strategy, start by calculating the number of moles of [Au(CN) 2 ] − present in the solution from the volume and concentration of the [Au(CN) 2 ] − solution:

\( \begin{align} moles\: [Au(CN)_2 ]^- & = V_L M_{mol/L} \\ & = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \end{align} \)

B Because the coefficients of gold and the [Au(CN) 2 ] − ion are the same in the balanced chemical equation, assuming that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN) 2 ] − (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so the number of moles of gold must be converted to the corresponding mass using the molar mass of gold:

\( \begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \\ &= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \end{align}\)

At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170.

\( 26 .0\: \cancel{g\: Au} \times \dfrac{1\: \cancel{troy\: oz}} {31 .10\: \cancel{g}} \times \dfrac{\$1400} {1\: \cancel{troy\: oz\: Au}} = \$1170 \)

Exercise \(\PageIndex{2}\) : Lanthanum Oxalate

What mass of solid lanthanum(III) oxalate nonahydrate [La 2 (C 2 O 4 ) 3 •9H 2 O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl 3 by adding a stoichiometric amount of sodium oxalate?

Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]

Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.

stoichiometry solved problems pdf

Solving Stoichiometry Problems

stoichiometry example problems

Core Concepts

In this tutorial, you will learn what is stoichiometry and the different types of problems involving it. You will go through several examples to practice and master the content!

Topics Covered in Other Articles

Balancing chemical equations.

  • What is a Chemical Reaction?
  • Calculating Molar Mass
  • Percent Yield Calculation
  • How to Calculate Molarity
  • What Are Significant Figures

Stoichiometry Definition

What is stoichiometry.

Stoichiometry is math having to do with chemical reactions. There are different types of calculations you can perform; stoichiometry with moles is the most common, but you can also do math with masses and even percentages. Read about the origins of stoichiometry here ! Learn what is mole in chemistry .

Stoichiometric Ratio

A stoichiometric ratio comes into play when talking about the relationships of elements or molecules in specific problems. This is the exact ratio between the coefficients of the reactants and products needed for a reaction to proceed normally. Let’s work through some problems you may see when learning about stoichiometry.

image illustrating stoichiometry and stoichiometric ratio

Stoichiometry Problems

A very common type of stoichiometric problem is balancing equations. This is an important chemistry skill to have because you have to have the correct ratio of reactants and products in order for a reaction to proceed; this is also an important foundation for organic chemistry. Although we have a tutorial on balancing equations , let’s look at one example.

Balance the following reaction:

\begin{gather*} {\_\_ C_{2}H_{2} + \_\_ O_{2} \rightarrow \_\_ CO_{2} + \_\_ H_{2}O} \end{gather*}

The main idea when balancing equations is that there should be the same number of each element on both sides of the reaction. You can balance the carbons and the hydrogens first, then move onto the oxygen. The balanced equation should look like this:

\begin{gather*} {2C_{2}H_{2} + 5O_{2} \rightarrow 4CO_{2} + 2H_{2}O} \end{gather*}

Example – Using Stoichiometric Ratio (Moles)

Use the equation below to solve the problem.

\begin{gather*} {C_{6}H_{12}O_{6} \rightarrow 2C_{2}H_{5}OH + 2CO_{2}} \end{gather*}

Example – Using Stoichiometric Ratio (Mass)

Let’s use the same equation as the problem above.

10\text{g}

Similar to the previous problem, by using the stoichiometric ratio of reactant to product, you can find the answer. Dimensional analysis is used to go from grams of C 2 H 5 OH to molar mass to mole (stoichiometric) ratio and back to grams.

\begin{gather*} {10\text{g}C_{2}H_{5}OH \cdot \frac{1\text{mol}C_{2}H_{5}OH}{46.068\text{g}C_{2}H_{5}OH} \cdot  \frac{1\text{mol}C_{6}H_{12}O_{6}}{2\text{mol}C_{2}H_{5}OH} \cdot \frac{180.156\text{g}C_{6}H_{12}O_{6}}{1\text{mol}C_{6}H_{12}O_{6}}  = 19.55\text{g}C_{6}H_{12}O_{6}} \end{gather*}

Stoichiometry Practice Problems

2.00\text{mol}

CaCl 2 and AgNO 3 react according to the following equation:

\begin{gather*} {CaCl_{2}(aq) + 2AgNO_{3}(aq) \rightarrow Ca\left(NO_{3}\right)_{2}(aq) + 2AgCl(s)} \end{gather*}

Hexane combusts according to the following reaction:

\begin{gather*} {2C_{6}H_{14}(g) + 19O_{2}(g) \rightarrow 12CO_{2}(g) + 14H_{2}O(g)} \end{gather*}

Stoichiometry Practice Problem Solutions

573\text{g}AgCl

For more help, view our interactive lecture on introducing stoichiometry problems!

And this video explaining how to solve reaction stoichiometry problems, further reading.

  • Percent by Weight Calculation
  • Balancing Redox Reactions
  • How to Write Net Ionic Equations
  • What is Specific Heat?
  • Common Ion Effect

Chemistry Steps

Chemistry Steps

stoichiometry solved problems pdf

General Chemistry

Stoichiometry.

This is a comprehensive, end-of-chapter set of practice problems on stoichiometry that covers balancing chemical equations, mole-ratio calculations, limiting reactants, and percent yield concepts. 

The links to the corresponding topics are given below.

  • The Mole and Molar Mass
  • Molar Calculations
  • Percent Composition and Empirical Formula
  • Stoichiometry of Chemical Reactions

Limiting Reactant

  • Reaction/Percent Yield
  • Stoichiometry Practice Problems

Balance the following chemical equations:

a) HCl + O 2 → H 2 O + Cl 2

b) Al(NO 3 ) 3 + NaOH → Al(OH) 3 + NaNO 3

c) H 2 + N 2 → NH 3

d) PCl 5 + H 2 O → H 3 PO 4 + HCl

e) Fe + H 2 SO 4 → Fe 2 (SO 4 ) 3 + H 2

f) CaCl 2 + HNO 3 → Ca(NO 3 ) 2 + HCl

g) KO 2 + H 2 O → KOH + O 2 + H 2 O 2

h) Al + H 2 O → Al 2 O 3 + H 2

i) Fe + Br 2 → FeBr 3

j) Cu + HNO 3 → Cu(NO 3 ) 2 + NO 2 + H 2 O

k) Al(OH) 3 → Al 2 O 3 + H 2 O

l) NH 3 + O 2 → NO + H 2 O

m) Ca(AlO 2 ) 2 + HCl → AlCl 3 + CaCl 2 + H 2 O

n) C 5 H 12 + O 2 → CO 2 + H 2 O

o) P 4 O 10 + H 2 O → H 3 PO 4

p) Na 2 CrO 4 + Pb(NO 3 ) 2 → PbCrO 4 + NaNO 3

q) MgCl 2 + AgNO 3 → AgCl + Mg(NO 3 ) 2

r) KClO 3 → KClO 4 + KCl

s) Ca(OH) 2 + H 3 PO 4 → Ca 3 (PO 4 ) 2 + H 2 O

Consider the balanced equation:

C 5 H 12 + 8 O 2 → 5CO 2 + 6H 2 O

Complete the table showing the appropriate number of moles of reactants and products.

How many grams of CO 2  and H 2 O are produced from the combustion of 220. g of propane (C 3 H 8 )?

C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g)

How many grams of CaCl 2 can be produced from 65.0 g of Ca(OH) 2 according to the following reaction,

Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O

How many moles of oxygen are formed when 75.0 g of Cu(NO 3 ) 2 decomposes according to the following reaction?

2Cu(NO 3 ) 2   → 2CuO + 4NO 2  + O 2

How many grams of MnCl 2  can be prepared from 52.1 grams of MnO 2 ?

MnO 2  + 4HCl → MnCl 2  + Cl 2  + 2H 2 O

Determine the mass of oxygen that is formed when an 18.3-g sample of potassium chlorate is decomposed according to the following equation:

2KClO 3 (s) → 2KCl(s) + 3O 2 (g).

How many grams of H 2 O will be formed when 48.0 grams H 2 are mixed with excess hydrogen gas?

2H 2  + O 2 → 2H 2 O

Consider the chlorination reaction of methane (CH4):

CH 4 (g) + 4Cl 2 (g) → CCl 4 (g) + 4HCl(g)

How many moles of CH 4 were used in the reaction if 51.9 g of CCl4 were obtained?

How many grams of Ba(NO 3 ) 2 can be produced by reacting 16.5 g of HNO 3 with an excess of Ba(OH) 2 ?

Ethanol can be obtained by fermentation – a complex chemical process breaking down glucose to ethanol and carbon dioxide.

                                                  C 6 H 12 O 6    →    2C 2 H 5 OH   +    2CO 2                                                       glucose                   ethanol

How many mL of ethanol (d =0.789 g/mL) can be obtained by this process starting with 286 g of glucose?

36.0 g of butane (C 4 H 10 ) was burned in an excess of oxygen and the resulting carbon dioxide (CO 2 ) was collected in a sealed vessel.

2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O

How many grams of LiOH will be necessary to consume all the CO 2 from the first reaction?

2LiOH + CO 2 → Li 2 CO 3 + H 2 O

13. Which statement about limiting reactant is correct?

a) The limiting reactant is the one in a smaller quantity.

b) The limiting reactant is the one in greater quantity.

c) The limiting reactant is the one producing less product.

d) The limiting reactant is the one producing more product.

Find the limiting reactant for each initial amount of reactants.

4NH 3 + 5O 2 → 4NO + 6H 2 O

a) 2 mol of NH 3 and 2 mol of O 2

b) 2 mol of NH 3 and 3 mol of O 2

c) 3 mol of NH 3 and 3 mol of O 2

d) 3 mol of NH 3 and 2 mol of O 2

Note:  This is not a multiple-choice question. Each row represents a separate question where you need to determine the limiting reactant.

How many g of hydrogen are left over in producing ammonia when 14.0 g of nitrogen is reacted with 8.0 g of hydrogen?

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

How many grams of PCl 3 will be produced if 130.5 g Cl 2 is reacted with 56.4 g P 4 according to the following equation?

6Cl 2 (g) + P 4 (s) → 4PCl 3 (l)

How many grams of sulfur can be obtained if 12.6 g H 2 S is reacted with 14.6 g SO 2 according to the following equation?

2H 2 S(g) + SO 2 (g) → 3S(s) + 2H 2 O(g)

The following equation represents the combustion of octane, C 8 H 18 , a component of gasoline:

2C 8 H 18 (g) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O(g)

Will 356 g of oxygen be enough for the complete combustion of 954 g of octane?

When 140.0 g of AgNO 3 was added to an aqueous solution of NaCl, 86.0 g of AgCl was collected as a white precipitate. Which salt was the limiting reactant in this reaction? How many grams of NaCl were present in the solution when AgNO 3 was added?

AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq)

Consider the reaction between MnO 2 and HCl:

MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O

What is the theoretical yield of MnCl 2 in grams when 165 g of MnO 2 is added to a solution containing 94.2 g of HCl?

Percent Yield

21. In a chemistry experiment, a student obtained 5.68 g of a product. What is the percent yield of the product if the theoretical yield was 7.12 g?

When 38.45 g CCl 4 is reacted with an excess of HF, 21.3 g CCl 2 F 2 is obtained. Calculate the theoretical and percent yields of this reaction.

CCl 4 + 2HF → CCl 2 F 2 + 2HCl

Iron(III) oxide reacts with carbon monoxide according to the equation:

Fe 2 O 3 ( s ) + 3CO( g ) → 2Fe( s ) + 3CO 2 ( g )

What is the percent yield of this reaction if 623 g of iron oxide produces 341 g of iron?

Determine the percent yield of the reaction if 77.0 g of CO 2  are formed from burning 2.00 moles of C 5 H 12  in 4.00 moles of O 2 .

C 5 H 12 + 8 O 2 → 5CO 2  + 6H 2 O

The percent yield for the following reaction was determined to be 84%:

N 2 ( g ) + 2H 2 ( g ) → N 2 H 4 ( l )

How many grams of hydrazine (N 2 H 4 ) can be produced when 38.36 g of nitrogen reacts with 6.68 g of hydrogen?

Silver metal can be prepared by reducing its nitrate, AgNO 3  with copper according to the following equation:

Cu( s ) + 2AgNO 3 ( aq ) → Cu(NO 3 ) 2 ( aq ) + 2Ag( s )

What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 132.5 grams of AgNO 3  ?

Industrially, nitric acid is produced from ammonia by the Ostwald process in a series of reactions:

4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2 O( l )

2NO( g ) + O 2 ( g ) → 2NO 2 ( g )

2NO 2 ( g ) + H 2 O( l ) → HNO 3 ( aq ) + HNO 2 ( aq )

Considering that each reaction has an 85% percent yield, how many grams of NH 3 must be used to produce 25.0 kg of HNO 3 by the above procedure?

Aspirin (acetylsalicylic acid) is widely used to treat pain, fever, and inflammation. It is produced from the reaction of salicylic acid with acetic anhydride. The chemical equation for aspirin synthesis is shown below:

stoichiometry solved problems pdf

In one container, 10.00 kg of salicylic acid is mixed with 10.00 kg of acetic anhydride.

a)  Which reactant is limiting? Which is in excess? b)  What mass of excess reactant is left over? c)  What mass of aspirin is formed assuming 100% yield (Theoretical yield)? d)  What mass of aspirin is formed if the reaction yield is 70.0% ? e)  If the actual yield of aspirin is 11.2 kg, what is the percent yield? f)  How many kg of salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?

3 thoughts on “Stoichiometry Practice Problems”

You forgot the subscript 3 for O in the molecular formula for acetic anhydride and the reaction is not balanced as written. For part F) it’s 18.1 kg and not1.81 kg as written in the final line of the solution.

Thanks for letting me know! Fixed.

You’re welcome!

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  3. Stoichiometry Gizmo Teacher Guide Answer Key: Your Ultimate Solution

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  5. Regular Chemistry

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VIDEO

  1. Lecture 09: Stoichiometry of Microbial Growth-I

  2. Stoichiometry Practice Problems

  3. Using Stoichiometry to Solve Electrolysis Problems (Updated)

  4. What is Stoichiometry? ll (FSC 1, Chapter 1) MDCAT Chemistry

  5. Ch2 ( stoichiometric calculation )

  6. Stoichiometric Calculations

COMMENTS

  1. PDF Stoichiometry: Problem Sheet 1

    Stoichiometry: Problem Sheet 1 Name: ________________________ Hour: ____ Date: ___________ acid react according to the following balanced equation: B. From the amount of nitric acid given in Part A, how many moles of silver nitrate will be produced? C. From the amount of nitric acid given in Part A, how many moles of water will be produced? D.

  2. PDF Stoichiometry with Solutions Problems

    Stoichiometry with Solutions Name ____________________ 1. H3PO4 + 3 NaOH --> Na3PO4 + 3 H2O How much 0.20 M H3PO4 is needed to react with 100 ml. of 0.10 M NaOH? 2. 2 HCl + Zn --> ZnCl2 + H2 When you use 25 ml. of 4.0 M HCl to produce H2 gas, how many grams of zinc does it react with? What volume of H2 gas is produced at STP? 3.

  3. 5.2.1: Practice Problems- Reaction Stoichiometry

    Answer. PROBLEM 5.2.1.5 5.2.1. 5. Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO 2, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction.

  4. PDF Chapter 3 Stoichiometry

    Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH4 (g) + 2O2 (g) CO2 (g) + 2 H2O (g) Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants appear on the left side of the equation. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

  5. PDF Stoichiometry

    moles etc - in the left-hand column; Step 3: insert all known information. Remember you will always be able to determine A r, M r, reaction coefficients and undertake a mass balance; . Step 4: use question mark (s) to identify the parameters you need to calculate to solve the problem; Step 5: identify which gaps in the framework you can calculate using the known or

  6. PDF STOICHIOMETRY AND MOLE CONCEPT

    STOICHIOMETRY AND MOLE CONCEPT SUMMARY NOTES & PRELIM QUESTIONS CONCEPT 1 BASIC MOLE CONCEPT CALCULATIONS Number of mole, number of particles = = Avogadro's constant, i.e. 6×1023 Number of mole, mass (g) mol-1) = = molar mass (g Number of mole, volume of gas = = molar volume of gas, i.e. 24 dm3 at r.t.p. Number of mole,

  7. PDF CHM 130 Stoichiometry Worksheet

    The following flow chart may help you work stoichiometry problems. Remember to pay careful attention to what you are given, and what you are trying to find. Fermentation is a complex chemical process of making wine by converting glucose into ethanol and carbon dioxide: C6H12O6(s) 2 C2H5OH (l) + 2 CO2(g)

  8. PDF Chapter 13 Stoichiometry

    Stoichiometry (STOY-key-OM-etry) problems are based on quantitative relationships between the different substances involved in a chemical reaction. 13.1 Mole Ratio The coefficients in a balanced equation given the moles of each substance in that equation.

  9. PDF Stoichiometry: Problem Sheet 2

    Directions: Solve each of the following problems. Show your work, including proper units, to earn full credit. ___ CaCl2 + ___ AgNO3 ___ Ca(NO3)2 + ___ AgCl How many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate? mol CaCl 2 mol AgCl 143 .5 g AgCl g AgCl 45 g CaCl 2

  10. PDF Stoichiometrty

    Classwork Set 1: 1) 2C2H6 + 7O2 --> 4CO2 + 6H2O a) How many moles of O2 are required to react with 24 moles of C2H6? b) How many grams of C2H6 are required to react with 12 moles of O2? c) How many grams of O2 are required to react with 200g of C2H6? 2) 2KClO3 --> 2KCl + 3O2 a) How many moles of O2 are required to react with 19 moles of KClO3?

  11. PDF STOICHIOMETRY PROBLEMS

    step of this strategy and then combine these steps for more complicated problems. Converting Quantity A to Mols A The more ways you can find the mols of a substance, the easier stoichiometry problems will become. Many times the units will help you get to your goal. Take for instance converting the mass of a substance to moles of the substance ...

  12. PDF Skills Worksheet Problem Solving

    General Plan for Solving Stoichiometry Problems 1 Mass of substance A Convert using the molar mass of A. 2 Amount in mol of substance A 3 Amount in mol of substance B Convert using the mole ratio A, given in the B balanced chemical equation. Convert using the molar mass of B. 4 Mass of substance B

  13. 3.5: Stoichiometry calculations with solutions

    The balanced chemical equation is as follows: 2FeCl 3(aq) + 3Na 2C 2O 4(aq) → Fe 2(C 2O 4) 3(s) + 6NaCl(aq) Solution. First we need to determine the number of moles of Na 2 C 2 O 4 that reacted. We will convert the volume to liters and then use the concentration of the solution as a conversion factor: 9.04 mL ×1 L1000.

  14. 5.3: Stoichiometry Calculations

    The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined above can be adapted as follows: 1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:

  15. PDF Lecture Notes: Stoichiometry

    problem. To summarize: when you solve a stoichiometry problem, first you must determine which substance is the limiting reagent. Many times this part of the problem is solved for you, making the problem a whole lot faster to solve. (After you figure out which substance is the limiting reagent, you should temporarily ignore the excess reagent ...

  16. PDF 12 Stoichiometry Stoichiome Planning Guide

    12.2.2 Explain the general procedure for solving a stoichiometric problem. Reading and Study Workbook Lesson 12.2 Lesson Assessment 12.2 p 398 Small-Scale Lab: Analysis of Baking Soda, p 399 ... step-by-step tutorials for solving various stoichiometry problems. Students can practice key problem-solving skills in an online problem set. T U O R M AT

  17. PDF Stoichiometry: Calculations with Chemical Formulas and Equations

    Percent Composition. One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: (number of atoms)(atomic weight) % element = (FW of the compound) x 100. Percent Composition. So the percentage of carbon in ethane is... (2)(12.0 amu)

  18. Stoichiometry (article)

    Step 1: Convert known reactant mass to moles. In order to relate the amounts H A 2 SO A 4 and NaOH using a mole ratio, we first need to know the quantity of H A 2 SO A 4 in moles. We can convert the 3.10 grams of H A 2 SO A 4 to moles using the molar mass of H A 2 SO A 4 ( 98.08 g / mol ):

  19. PDF Chapter 1: Stoichiometry

    1 Chapter 1: Stoichiometry 1. The atomic mass of C is 12.011 u. How many moles of C are there in a 3.50 g sample of carbon? * a. 0.291 moles

  20. Stoichiometry

    Stoichiometry Practice Problems. Problem 1. You mix of CaCl 2 into a solution with excess AgNO 3, hence CaCl 2 is your limiting reactant. Assuming a 100% yield, how much AgCl would you expect to produce, in grams? (AgCl has a molecular weight of ). CaCl 2 and AgNO 3 react according to the following equation:

  21. Stoichiometry Practice Problems

    answer This content is available to registered users only. Click here to Register! By joining Chemistry Steps, you will gain instant access to the Answers and Solutions for all the Practice Problems, Quizzes, and the powerful set of General Chemistry 1 and 2 Summary Study Guides.

  22. PDF Teaching Stoichiometry as Algebraic Word Problems, Part 2

    Stoichiometry is a basic topic of chemistry, concerned with solving for certain quantities of products and/or reactants in (usually) a balanced chemical equa-tion, given knowledge of other quantities of products and/or reactants in the equation. Such quantities of interests are typically moles, grams, and/or liters of particular substances.

  23. PDF Worksheet 4.6 Gas stoichiometry

    This worksheet provides practice in stoichiometry problems involving gases at standard conditions (STP) and non-standard conditions. The relevant formulas for calculations are: n = V m V and n = RT PV where V m = 22.4 dm3 mol-1 at STP (T = 0 C, P = 1 atm). The second half of the worksheet looks at problems relating to gas densities, where the ...