Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 9, Problem 89AP

For each of the following unbalanced chemical equations, suppose 25.0 g of each reactant is taken. Show by calculation which reactant is limiting. Calculate the theoretical yield in grams of the product in boldface.

msp;  C 2 H 5 OH ( l ) + O 2 ( g ) CO 2 ( g ) + H 2 O ( l )

msp;  N 2 ( g ) + O 2 ( g ) NO ( g )

msp;  NaClO 2 ( a q ) + Cl 2 ( g ) NaCl ( a q )

msp;  H 2 ( g ) + N 2 ( g ) NH 3 ( g )

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

To determine the limiting and calculate the theoretical yield of product in given reaction

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Answer to Problem 89AP

In this reaction O2 is a limiting reactant and there are 22.96 g CO2 formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

C2H5OH(l) + 3O2(g)Δ 2CO2(g) + 3 H2O(g) 

Given:

Amount of C2H5OH = 25.0 g

Amount of O2 = 25.0 g

Calculation:

Number of moles of C2H5OH and O2 calculated as follows:

Number of moles=mass in gmolarmass=25.0 g46.07 g/mol=0.542 moles C2H5OHNumber of moles=mass in gmolarmass=25.0 g32.00 g/mol=0.781 moles O2

In this reaction O2 is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

0.781 moles O2×2.00 mole CO23.00 moles O2×44.1 g CO21.00 mole CO2=22.96 g CO2.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

To determine the limiting and calculate the theoretical yield of product in given reaction

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Answer to Problem 89AP

In this reaction O2 is a limiting reactant and there are 23.43 g NO formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

N2(g) + O2(g) 2NO(g)

Given:

Amount of N2 = 25.0 g

Amount of O2 = 25.0 g

Calculation:

Number of moles of N2 and O2 calculated as follows:

Number of moles=mass in gmolarmass=25.0 g28.013 g/mol=0.892 moles N2Number of moles=mass in gmolarmass=25.0 g32.00 g/mol=0.781 moles O2

In this reaction O2 is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

0.781 moles O2×1.00 mole NO1.00 moles O2×30.01 g NO1.00 mole NO=23.43 g NO.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

To determine the limiting and calculate the theoretical yield of product in given reaction

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Answer to Problem 89AP

In this reaction NaClO2 is a limiting reactant and there are 16.13 g NaCl formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

2NaClO2(aq) + Cl2(g) 2ClO2(g)+2NaCl(aq)

Given:

Amount of NaClO2 = 25.0 g

Amount of Cl2 = 25.0 g

Calculation:

Number of moles of NaClO2 and Cl2 calculated as follows:

Number of moles=mass in gmolarmass=25.0 g90.44 g/mol=0.276 moles NaClO2Number of moles=mass in gmolarmass=25.0 g70.906 g/mol=0.353 moles Cl2

In this reaction NaClO2 is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

0.276 moles NaClO2×2.00 mole NaCl2.00 moles NaClO2×58.44 g NaCl1.00 mole NaCl=16.13 g NaCl.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

To determine the limiting and calculate the theoretical yield of product in given reaction

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Answer to Problem 89AP

In this reaction N2 is a limiting reactant and there are 30.38 g NH3 formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

3H2(aq) + N2(g) 2NH3(g)

Given:

Amount of H2 = 25.0 g

Amount of N2 = 25.0 g

Calculation:

Number of moles of H2 and N2 calculated as follows:

Number of moles=mass in gmolarmass=25.0 g2.016 g/mol=12.4 molesH2Number of moles=mass in gmolarmass=25.0 g28.013 g/mol=0.892 moles N2

In this reaction N2 is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

0.892 moles N2×2.00 mole NH31.00 moles N2×17.031 g NH31.00 mole NH3=30.38 g NH3.

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Chapter 9 Solutions

Introductory Chemistry: A Foundation

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