EXAMPLE 8.1

Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Introductory Chemistry, Third EditionBy Nivaldo J. Tro

You are given the number of moles of a reactant (Cl2) and asked to find how many moles of product (NaCl) will result if the reactant completely reacts. The conversion factor comes from the balanced chemical equation.

EXAMPLE 8.1 Mole-to-Mole Conversions

The solution map begins with moles of chlorine and uses the stoichiometric conversion factor to obtain moles of sodium chloride

Follow the solution map to solve the problem. There is enough Cl2 to produce 6.8 mol of NaCl.

Solution Map:

Sodium chloride, NaCl, forms by the following reaction between sodium and chlorine.

How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2 ? Assume that there is more than enough Na.

Solution:





FOR MORE PRACTICE Example 8.8; Problems 17, 18, 19, 20.

Continued

Water is formed when hydrogen gas reacts explosively with oxygen gas according to the following balanced

equation.

How many moles of H2O result from the complete reaction of 24.6 mol of O2? Assume that there is more than

Enough H2.

SKILLBUILDER 8.1 Mole-to-Mole Conversions




Set up the problem in the normal way.

The main conversion factor is the stoichiometric relationship between moles of carbon dioxide and moles of glucose. This conversion factor comes from the balanced equation. The other conversion factors are simply the molar masses of carbon dioxide and glucose.

EXAMPLE 8.2 Mass-to-Mass Conversions

In photosynthesis, plants convert carbon dioxide and water into glucose (C6 H12O6) according to the following reaction.

How many grams of glucose can be synthesized from 58.5 g of CO2 ? Assume that there is more than enough water present to react with all of the CO2 .

The solution map uses the general outline

where A is carbon dioxide and B is glucose.

Solution Map:





FOR MORE PRACTICE Example 8.9; Problems 33, 34, 35, 36.

Continued

Follow the solution map to solve the problem. Begin with grams of carbon dioxide and multiply by the appropriate factors to arrive at grams of glucose.

Solution:

Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl, according to the following reaction.

How much HCl in grams can be neutralized by 5.50 g of Mg(OH)2?

SKILLBUILDER 8.2 Mass-to-Mass Conversions




Set up the problem in the normal way.

The main conversion factor is the stoichiometric relationship between moles of nitrogen dioxide and moles of nitric acid. This conversion factor comes from the balanced equation. The other conversion factors are simply the molar masses of nitrogen dioxide and nitric acid and the relationship between kilograms and grams.


One of the components of acid rain is nitric acid, which forms when NO2 a pollutant, reacts with oxygen and rainwater according to the following reaction.

Assuming that there is more than enough O2 and H2O, how much HNO3 in kilograms forms from 1.5 103 kg of NO2 pollutant?

Solution Map:The solution map follows the general format of:





FOR MORE PRACTICE Problems 37, 38, 39, 40.

Continued

However, since the original quantity of NO2 is given in kilograms, you must first convert to grams. Since the final quantity is requested in kilograms, you must convert back to kilograms at the end.Follow the solution map to solve the problem. Begin with kilograms of nitrogen dioxide and multiply by the appropriate conversion factors to arrive at kilograms of nitric acid.

Solution Map:

Solution:

Another component of acid rain is sulfuric acid, which forms when SO2 also a pollutant, reacts with oxygen and rainwater according to the following reaction.

Assuming that there is plenty of O2 and H2O how much H2SO4 in kilograms forms from 2.6 × 103 kg of SO2?

SKILLBUILDER 8.3 Mass-to-Mass Conversions




Set up the problem in the normal way. The conversion factors are the stoichiometric relationships (from the balanced equation) between the moles of each reactant and the moles of product.

EXAMPLE 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants

The solution map shows how to get from moles of each reactant to moles of AlCl3. The reactant that makes the least amount of AlCl3 is the limiting reactant..

Consider the following reaction.

If we begin with 0.552 mol of aluminum and 0.887 mol of chlorine, what is the limiting reactant and theoretical yield of AlCl3 in moles?

Solution Map:

Given: 0.552 mol Al0.887 mol Cl2

Find: limiting reactanttheoretical yield




EXAMPLE 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants

FOR MORE PRACTICE Problems 45, 46, 47, 48, 49, 50, 51, 52.

Continued

Solution:


If you begin with 4.8 mol of sodium and 2.6 mol of fluorine, what is the limiting reactant and theoretical yield of NaF in moles?

SKILLBUILDER 8.4 Limiting Reactant and Theoretical Yield from Initial Moles of Reactants

Since the 0.552 mol of Al makes the least amount of AlCl3 is the limiting reactant. The theoretical yield is 0.552 mol of AlCl3.




Although this problem does not specifically ask for the limiting reactant, it must be found to determine the theoretical yield, which is the maximum amount of ammonia that can be synthesized. Begin by setting up the problem in the normal way. The main conversion factors are the stoichiometric relationship between moles of each reactant and moles of ammonia. The other conversion factors are simply the molar masses of nitrogen monoxide, hydrogen gas, and ammonia.

EXAMPLE 8.5 Finding Limiting Reactant and Theoretical Yield

Given: 545.8 g NO, 12.4 g H2

Find: maximum amount of NH3 (theoretical yield)Conversion Factors:

Ammonia, NH3, can be synthesized by the following reaction.

What is the maximum amount of ammonia in grams that can be synthesized from 45.8 g of NO and 12.4 g of H2?





Continued

Find the limiting reactant by calculating how much product can be made from each reactant. The reactant that makes the least amount of product is the limiting reactant.

Solution Map:





Continued

Follow the solution map, beginning with the actual amount of each reactant given, to calculate the amount of product that can be made from each reactant.

There is enough NO to make 26.0 g of NH3 and enough H2 to make 41.8 g of HN3. Therefore, NO is the limitingreactant, and the maximum amount of ammonia that can possibly be made is 26.0 g, the theoretical yield.

Solution:

Ammonia can also be synthesized by the following reaction.

What is the maximum amount of ammonia in grams that can be synthesized from 25.2 g of N2 and 8.42 g of H2?

SKILLBUILDER 8.5 Finding Limiting Reactant and Theoretical Yield





Continued

SKILLBUILDER PLUS

What is the maximum amount of ammonia in kilograms that can be synthesized from 5.22 kg of H2 and 31.5 kg of N2?

FOR MORE PRACTICE Problems 55, 56, 57, 58.




Set up the problem in the normalway.

The main conversion factors are the stoichiometric relationships between moles of each reactant and moles of copper. The other conversion factors are simply the molar masses of copper(I) oxide, carbon, and copper

EXAMPLE 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield


When 11.5 g of C are allowed to react with 114.5 g of Cu2O, 87.4 g of Cu, are obtained. Find the limiting reactant,theoretical yield, and percent yield.

Given: 11.5 g C114.5 g Cu2O87.4 g Cu produced

Find: limiting reactanttheoretical yieldpercent yield





Continued

The solution map shows how to find the mass of Cu formed by the initial masses of Cu2O and C. The reactant that makes the least amount of product is the limiting reactant and determines the theoretical yield.

Solution Map:

Follow the solution map, beginning with the actual amount of each reactant given, to calculate the amount of product that can be made from each reactant.

Since Cu2O makes the least amount of product, Cu2O is the limiting reactant. The theoretical yield is simply the amount of product made by the limiting reactant. The percent yield is the actual yield (87.4 g Cu) divided by the theoretical yield (101.7 g Cu) multiplied by 100%.

Solution:





Continued

FOR MORE PRACTICE Example 8.10; Problems 61, 62, 63, 64, 65, 66.

The following reaction is used to obtain iron from iron ore:

The reaction of 185 g of Fe2O3 with 95.3 g of CO produces 87.4 g of Fe. Find the limiting reactant, theoretical yield, and percent yield.

SKILLBUILDER 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield




You are given the mass of propane and asked to find the heat evolved (in kJ) in its combustion.

EXAMPLE 8.7 Stoichiometry Involving ΔH

Start with g C3H8 and then use the molar mass of C3H8 to find the number of moles. Next, use the stoichiometric relationship between mol C3H8 and kJ to find the heat evolved.

Solution Map:

An LP gas tank in a home barbecue contains 11.8 × 103 g of propane (C3H8). Calculate the heat (in kJ) associatedwith the complete combustion of all of the propane in the tank.

Given: 11.8 × 103 g C3H8

Find: kJConversion Factors:





FOR MORE PRACTICE Example 8.11; Problems 69, 70, 71, 72, 73, 74.

Continued

Follow the solution map to solve the problem. Begin with 11.8 × 103 g C3H8 andmultiply by the appropriate conversion factors to arrive at kJ. The answer is negative, as it should be for heat evolved by the reaction.

Solution:

Ammonia reacts with oxygen according to the following equation:

Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH3.

SKILLBUILDER 8.7 Stoichiometry Involving ΔH

What mass of butane in grams is necessary to produce 1.5 × 103 kJ of heat? What mass of CO2 is produced?

SKILLBUILDER PLUS





How many moles of sodium oxide can be synthesized from 4.8 mol of sodium? Assume that there is more than enough oxygen present. The balanced equation is:

Given: 4.8 mol Na

Find: mol Na2O

Conversion Factor:

Solution Map:

Solution:





How many grams of sodium oxide can be synthesized from 17.4 g of sodium? Assume that there is more than enough oxygen present. The balanced equation is:

Given: 17.4 g Na

Find: g Na2O

Conversion Factors:

Molar mass Na = 22.99 g/molMolar mass Na2O = 61.98 g/mol

Solution Map:





Continued

Solution:




EXAMPLE 8.10 Limiting Reactant, Theoretical Yield, and Percent Yield

10.4 g of Fe reacts with 11.8 g of S to produce 14.2 g of Fe2S3. Find the limiting reactant, theoretical yield, and percent yield for this reaction. The balanced chemical equation is:

Given: 10.4 g Fe11.8 g S14.2 g Fe2S3

Find: limiting reactanttheoretical yieldpercent yield





Continued

Solution Map:





Continued

The limiting reactant is Fe.The theoretical yield is 19.4 g of Fe2S3 .

The percent yield is 73.2%.

Solution:





Calculate the heat evolved (in kJ) upon complete combustion of 25.0 g of CH4.

Given: 25 g CH4

Find: kJ

Conversion Factors:

Solution Map:

Solution:

Documents

EXAMPLE 8.1