example exercise 10.1 interpreting chemical equation ... exercise 10.1 interpreting chemical...

© 2011 Pearson Education, Inc.Introductory Chemistry: Concepts and Critical Thinking, 6th EditionCharles H. Corwin

Example Exercise 10.1 Interpreting Chemical Equation Calculations

The balanced chemical equation is

(a) The coefficients in the equation (1:2:1:2) indicate the ratio of moles as well as molecules.

(b) The coefficients in the equation (1:2:1:2) indicate the ratio of volumes of gases. If we express the volume in liters, we have

Solution

Given the chemical equation for the combustion of methane, CH4, balance the equation and interpret the coefficients in terms of (a) moles and (b) liters:

Given the chemical equation for the combustion of propane, C3H8, balance the equation and interpret the coefficients in terms of (a) moles and (b) milliliters:

Answers:

Practice Exercise

Which of the following is in the same ratio as the coefficients in a balanced equation: moles of gas, mass of gas, volume of gas?

Answer: See Appendix G.

Concept Exercise


Example Exercise 10.2 Mole–Mole Relationships

We select the unit factor that cancels the unit in the given value . Thus,Solution

Carbon monoxide is produced in a blast furnace by passing oxygen gas over hot coal. How many moles of oxygen react with 2.50 mol of carbon according to the balanced equation?

Unit Analysis Map

Strategy PlanStep 1: What unit is asked for in the answer?

Step 2: What given value is related to the answer?

Step 3: What unit factor(s) should we apply?

Using the balanced equation, we see that 2 molC = 1 mol O2.Thus, the mole ratio and the twounit factors are


Example Exercise 10.2 Mole–Mole Relationships

Iron is produced from iron ore in a blast furnace by passing carbon monoxide gas through molten iron(III) oxide. The balanced equation is

(a) How many moles of carbon monoxide react with 2.50 mol of Fe2O3?(b) How many moles of iron are produced from 2.50 mol of Fe2O3?

Answers: (a) 7.50 mol CO; (b) 5.00 mol Fe

Practice Exercise

How many unit factors are required to solve a mole–mole stoichiometry problem?Answer: See Appendix G.

Concept Exercise

Continued


Example Exercise 10.3 Classifying Stoichiometry Problems

After analyzing a problem for the unknown quantity and the relevant given value, we classify the type of problem.(a) The problem asks for grams of Zn (mass) that react to give 0.500 g of ZnCl2 (mass). This is

a mass–mass type of problem.(b) The problem asks for liters of H2 gas (volume) that react to yield 50.0 cm3 of HCl gas

(volume). This is a volume–volume type of problem.(c) The problem asks for kilograms of Fe (mass) that react to produce 50.0 mL of H2 gas

(volume). This is a mass–volume type of problem.

Solution

Classify the type of stoichiometry problem for each of the following:(a) How many grams of Zn metal react with hydrochloric acid to give 0.500 g of zinc chloride?(b) How many liters of H2 gas react with chlorine gas to yield 50.0 cm3 of hydrogen chloride gas?(c) How many kilograms of Fe react with sulfuric acid to produce 50.0 mL of hydrogen gas?


Example Exercise 10.3 Classifying Stoichiometry Problems

Classify the type of stoichiometry problem for each of the following:(a) How many grams of HgO decompose to give 0.500 L of oxygen gas at STP?(b) How many grams of AgCl are produced from the reaction of 0.500 g of solid sodium chloride with

silver nitrate solution?(c) How many milliliters of H2 gas react with nitrogen gas to yield 1.00 L of ammonia gas?

Answers: (a) mass–volume; (b) mass–mass; (c) volume–volume

Practice Exercise

What are the three types of stoichiometry problems?Answer: See Appendix G.

Concept Exercise

Continued


Example Exercise 10.4 Mass–Mass Stoichiometry

We select from each of the three pairs of ratios a unit factor that cancels the previous units. Thus,Solution

Calculate the mass of mercury produced from the decomposition of 1.25 g of orange mercury(II) oxide (MM = 216.59 g/mol):

Unit Analysis Map

Strategy PlanStep 1: What unit is asked for in the answer?Step 2: What given value is related to the answer?Step 3: What unit factor(s) should we apply?We are given that 1 mol HgO = 216.59 g HgO; the mole ratio from the balanced equation is 2 mol HgO = 2 mol Hg. From the periodic table, we find that 1 mol Hg = 200.59 g Hg. Thus, the unit factors are



Calculate the mass of carbon dioxide released from 10.0 g of cobalt(III) carbonate given the unbalancedequation for the reaction:

Answer: 4.43 g CO2

Practice Exercise

What are the three steps in the unit analysis method of problem solving?Answer: See Appendix G.

Concept Exercise

Continued


Example Exercise 10.5 Mass–Mass StoichiometryCalculate the mass of potassium iodide (166.00 g/mol) required to yield 1.78 g of mercury(II) iodide precipitate (454.39 g/mol):

Unit Analysis Map



Step 3: What unit factor(s) should we apply?

From the given molar mass, we see that HgI2 = 454.39 g HgI2. The mole ratio from the balanced equation is KI = 1 mol HgI2, and from the given molar mass, 1 mol KI = 166.00 g KI. Thus, the three pairs of unit factors are



Calculate the mass of iron filings required to produce 0.455 g of silver metal given the unbalanced equation for the reaction:

Answer: 0.0785 g Fe

Practice Exercise

In general, how many unit factors are required to solve a mass–mass stoichiometry problem?


Concept Exercise


Continued


Example Exercise 10.6 Mass–Volume Stoichiometry


In an automobile collision, sodium azide, NaN3, decomposes and fills an air bag with nitrogen gas. If an air bag contains 100.0 g of NaN3 (65.02 g/mol), what is the volume of nitrogen gas produced at STP? The equation for the reaction is

Unit Analysis Map



Step 3: What unit factor(s) should we apply?Given the molar mass, NaN3 = 65.02 g NaN3. The moleratio from the balanced equation is 2 mol NaN3 = 3 mol N2,and the molar volume of a gas is mol N2 = 22.4 L N2.Thus, the unit factors are



Calculate the volume of hydrogen gas produced at STP from 1.55 g of sodium metal in water given the unbalanced equation for the reaction:

Answer: 0.755 L H2

Practice Exercise

What are the three steps in the unit analysis method of problem solving?


Concept Exercise

Continued




Baking soda can be used as a fire extinguisher. When heated, it decomposes to carbon dioxide gas, which can smother a fire. If a sample of NaHCO3 (84.01 g/mol) produces 0.500 L of carbon dioxide gas at STP, what is the mass of the sample?

Unit Analysis Map



Step 3: What unit factor(s) should we apply?From molar volume, we know that 1 mol CO2 = 22.4 L CO2. The mole ratio from the balanced equation is2 mol NaHCO3 = 1 mol CO2, and from the given molarmass, 1 mol NaHCO3 = 84.01 g NaHCO3.Thus, the unit factors are



Calculate the mass of aluminum metal required to release 2160 mL of hydrogen gas at STP from sulfuric acid given the unbalanced equation for the reaction:

Answer: 1.73 g Al

Practice Exercise

In general, how many unit factors are required to solve a mass–volume stoichiometry problem?


Concept Exercise

Continued


Example Exercise 10.8 Volume–Volume StoichiometryIn the Haber process, nitrogen and hydrogen gases combine to give ammonia gas. If 5.55 L of nitrogen gas is available, calculate the volume of ammonia that is produced. Assume that all volumes of gas are measured under constant conditions of 500 °C and 300 atm pressure:

Unit Analysis Map



Step 3: What unit factor(s) should we apply?Using the balanced equation, we see that molN2 = 2 mol NH3. Since the volumes of gas havethe same ratio as the mole ratio, the unit factors are


Example Exercise 10.8 Volume–Volume Stoichiometry

We select the unit factor that cancels the unit in the given value (L N2). Thus,

Solution

Calculate the volumes of (a) hydrogen chloride gas and (b) oxygen gas that react to yield 50.00 mL of chlorine gas given the following unbalanced equation. Assume that all gases are at the same temperature and pressure:

Answers: (a) 100.0 mL HCl; (b) 25.00 mL O2

Practice Exercise

In general, how many unit factors are required to solve a volume–volume stoichiometry problem?Answer: See Appendix G.

Concept Exercise

Continued


Example Exercise 10.9 Limiting Reactant ConceptA 1.00 mol sample of iron(II) oxide is heated with 1.00 mol of aluminum metal and converted to molten iron. Identify the limiting reactant, and calculate the moles of iron produced:

Unit Analysis Map



Step 3: What unit factor(s) should we apply?From the balanced equation, we have the mole ratio3 mol FeO/3 mol Fe, and 2 mol Al/3 mol Fe.


Example Exercise 10.9 Limiting Reactant Concept

We select the unit factor that cancels the unit in the given value . Thus,

Second, we select the unit factor that cancels the unit in the given value . Thus,

Notice that FeO produces 1.00 mol Fe, whereas Al produces 1.50 mol Fe. Thus, FeO is the limiting reactantand the amount of product is 1.00 mol Fe.

Solution

A 5.00 mol sample of iron(III) oxide is heated with 5.00 mol aluminum metal and converted to molten iron. Identify the limiting reactant, and calculate the moles of iron produced given the unbalanced equation for the reaction:

Answer: The limiting reactant is Al, which produces 5.00 mol of Fe.

Practice Exercise

A Tour de France cyclist has 10 tires and 3 frames. How many complete bicycles can be assembled?


Concept Exercise

Continued


Example Exercise 10.10 Mass–Mass Limiting Reactant

Step 1: We calculate the mass of Mn produced from 50.0 g of MnO2. We can outline the solution as follows:

From the periodic table, we find the molar mass of MnO2 is 86.94 g/mol and Mn is 54.94 g/mol. The unit analysis solution to the problem is

Step 2: We calculate the mass of Mn produced from 25.0 g of Al. We can outline the solution as follows:

From the periodic table, we find that the molar mass of Al is 26.98 g/mol. Starting with 25.0 g of Al, the unit analysis solution is

Solution

In a reaction, 50.0 g of manganese(IV) oxide reacts with 25.0 g of aluminum. Identify the limiting reactant and calculate the mass of manganese metal produced. The equation for the reaction is


Example Exercise 10.10 Mass–Mass Limiting Reactant

Step 3: We compare the mass of product obtained from each of the reactants:

In this example, MnO2 is the limiting reactant because it yields less product. Al is the excess reactant and is not completely consumed in the reaction. Thus, the maximum amount of product from the reaction is 31.6 g of Mn.

Solution

In a reaction, 75.0 g of manganese(IV) oxide react with 30.0 g of aluminum. Identify the limiting reactant and calculate the mass of aluminum oxide produced:

Answer: The limiting reactant is Al, which gives 56.7 g of Al2O3.

Practice Exercise

Indicate the limiting reactant shown in the following illustration


Concept Exercise

Continued


Example Exercise 10.11 Volume–Volume Limiting ReactantIn oxyacetylene welding, acetylene reacts with oxygen to give carbon dioxide and water. If 25.0 mL of C2H2 reacts with 75.0 mL of O2, what is the limiting reactant? Assuming constant conditions, what is the volume of CO2produced?

Unit Analysis Map



Step 3: What unit factor(s) should we apply?From the balanced equation, we have the mole ratio so the volume ratio is 2 mL C2H2/4 mL CO2, and 5 mL O2/4 mL CO2.


Example Exercise 10.11 Volume–Volume Limiting Reactant

Ethane undergoes combustion to give carbon dioxide and water. If 10.0 L of C2H6 reacts with 25.0 L of O2, what is the limiting reactant? Assuming constant conditions, what is the volume of CO2 produced? The equation is

Answer: The limiting reactant is O2, which gives 14.3 L of CO2.

Practice Exercise

First, we find the amount of product that can be produced from the first reactant (mL C2H2). Thus,

Second, we find the amount of product that can be produced from the second reactant (mL O2):

Third, we compare the amount of product from each reactant and see that C2H2 produces 50.0 mL CO2, whereas O2 produces 60.0 mL CO2. Thus, C2H2 is the limiting reactant and the amount of product is limited to 50.0 mL CO2.

Solution

Continued


Example Exercise 10.11 Volume–Volume Limiting Reactant

Indicate the limiting reactant shown in the following illustration:


Concept Exercise

Continued


Example Exercise 10.12 Percent Yield

The percent yield is the ratio of the actual yield compared to the theoretical yield. In this experiment, the actual yield is 0.875 g and the theoretical yield is 0.988 g. The percent yield, therefore, is

The percent yield obtained by the student is 88.6%.

Solution

A student dissolves 1.50 g of copper(II) nitrate in water. After adding aqueous sodium carbonate solution, the student obtains 0.875 g of CuCO3 precipitate. If the theoretical yield is 0.988 g, what is the percent yield?

Ammonium nitrate is used in explosives and is produced by the reaction of ammonia, NH3, and nitric acid. The equation for the reaction is

If 15.0 kg of ammonia gives an actual yield of 65.3 kg of ammonium nitrate, what is the percent yield? The calculated yield of ammonium nitrate for the experiment is 70.5 kg.

Answer: 92.6%

Practice Exercise


Example Exercise 10.12 Percent Yield

Is it possible to have a percent yield greater than 100%?


Concept Exercise

Continued

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