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;  ${\text{C}}_2{\text{H}}_5\text{OH}\left(l\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$

msp;  ${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to \text{NO}\left(g\right)$

msp;  ${\text{NaClO}}_2\left(aq\right)+{\text{Cl}}_2\left(g\right)\to \text{NaCl}\left(aq\right)$

msp;  ${\text{H}}_2\left(g\right)+{\text{N}}_2\left(g\right)\to {\text{NH}}_3\left(g\right)$

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:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 89AP

In this reaction ${\text{O}}_2$ is a limiting reactant and there are $22.96\text{g}\text{C}{\text{O}}_2$ 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:

${\text{C}}_2{\text{H}}_5\text{O}\text{H}(l)+\text{ 3}{\text{O}}_2(g)\stackrel{\text{Δ}}{\to }\text{ 2}\text{C}{\text{O}}_2(g)+3{\text{H}}_2\text{O}(g)$

Given:

Amount of ${\text{C}}_2{\text{H}}_5\text{O}\text{H}$ = 25.0 g

Amount of ${\text{O}}_2$ = 25.0 g

Calculation:

Number of moles of ${\text{C}}_2{\text{H}}_5\text{O}\text{H}$ and ${\text{O}}_2$ calculated as follows:

$\begin{array}{l}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{46}\text{.07 }\text{g}/\text{m}\text{o}\text{l}}=0.542\text{m}\text{o}\text{l}\text{e}\text{s}{\text{C}}_2{\text{H}}_5\text{O}\text{H}\\ \text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{32}\text{.00 }\text{g}/\text{m}\text{o}\text{l}}=0.781\text{m}\text{o}\text{l}\text{e}\text{s}{\text{O}}_2\end{array}$

In this reaction ${\text{O}}_2$ is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

$0.781\text{m}\text{o}\text{l}\text{e}\text{s}{\text{O}}_2\times \frac{\cancel{2.00\text{m}\text{o}\text{l}\text{e}\text{C}{\text{O}}_2}}{3.00\text{m}\text{o}\text{l}\text{e}\text{s}{\text{O}}_2}\times \frac{\text{44}\text{.1 }\text{g}\text{C}{\text{O}}_2}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{C}{\text{O}}_2}}=22.96\text{g}\text{C}{\text{O}}_2$.

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:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 89AP

In this reaction ${\text{O}}_2$ is a limiting reactant and there are $23.43\text{g}\text{N}\text{O}$ 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:

${\text{N}}_2(g)+{\text{O}}_2(g)\to 2\text{N}\text{O}(g)$

Given:

Amount of ${\text{N}}_2$ = 25.0 g

Amount of ${\text{O}}_2$ = 25.0 g

Calculation:

Number of moles of ${\text{N}}_2$ and ${\text{O}}_2$ calculated as follows:

$\begin{array}{l}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{28}\text{.013 }\text{g}/\text{m}\text{o}\text{l}}=0.892\text{m}\text{o}\text{l}\text{e}\text{s}{\text{N}}_2\\ \text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{32}\text{.00 }\text{g}/\text{m}\text{o}\text{l}}=0.781\text{m}\text{o}\text{l}\text{e}\text{s}{\text{O}}_2\end{array}$

In this reaction ${\text{O}}_2$ is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

$0.781\text{m}\text{o}\text{l}\text{e}\text{s}{\text{O}}_2\times \frac{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{N}\text{O}}}{1.00\text{m}\text{o}\text{l}\text{e}\text{s}{\text{O}}_2}\times \frac{\text{30}\text{.01 }\text{g}\text{N}\text{O}}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{N}\text{O}}}=23.43\text{g}\text{N}\text{O}$.

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:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 89AP

In this reaction $\text{N}\text{a}\text{C}\text{l}{\text{O}}_2$ is a limiting reactant and there are $16.13\text{g}\text{N}\text{a}\text{C}\text{l}$ 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:

$2\text{N}\text{a}\text{C}\text{l}{\text{O}}_2(aq)+\text{C}{\text{l}}_2(g)\to \text{ 2}\text{C}\text{l}{\text{O}}_2(g)+2\text{N}\text{a}\text{C}\text{l}(aq)$

Given:

Amount of $\text{N}\text{a}\text{C}\text{l}{\text{O}}_2$ = 25.0 g

Amount of $\text{C}{\text{l}}_2$ = 25.0 g

Calculation:

Number of moles of $\text{N}\text{a}\text{C}\text{l}{\text{O}}_2$ and $\text{C}{\text{l}}_2$ calculated as follows:

$\begin{array}{l}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{90}\text{.44 }\text{g}/\text{m}\text{o}\text{l}}=0.276\text{m}\text{o}\text{l}\text{e}\text{s}\text{N}\text{a}\text{C}\text{l}{\text{O}}_2\\ \text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{70}\text{.906 }\text{g}/\text{m}\text{o}\text{l}}=0.353\text{m}\text{o}\text{l}\text{e}\text{s}\text{C}{\text{l}}_2\end{array}$

In this reaction $\text{N}\text{a}\text{C}\text{l}{\text{O}}_2$ is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

$0.276\text{m}\text{o}\text{l}\text{e}\text{s}\text{N}\text{a}\text{C}\text{l}{\text{O}}_2\times \frac{\cancel{2.00\text{m}\text{o}\text{l}\text{e}\text{N}\text{a}\text{C}\text{l}}}{2.00\text{m}\text{o}\text{l}\text{e}\text{s}\text{N}\text{a}\text{C}\text{l}{\text{O}}_2}\times \frac{\text{58}\text{.44 }\text{g}\text{N}\text{a}\text{C}\text{l}}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{N}\text{a}\text{C}\text{l}}}=16.13\text{g}\text{N}\text{a}\text{C}\text{l}$.

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:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 89AP

In this reaction ${\text{N}}_2$ is a limiting reactant and there are $30.38\text{g}\text{N}{\text{H}}_3$ 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:

$3{\text{H}}_2(aq)+{\text{N}}_2(g)\to \text{ 2}\text{N}{\text{H}}_3(g)$

Given:

Amount of ${\text{H}}_2$ = 25.0 g

Amount of ${\text{N}}_2$ = 25.0 g

Calculation:

Number of moles of ${\text{H}}_2$ and ${\text{N}}_2$ calculated as follows:

$\begin{array}{l}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{2}\text{.016 }\text{g}/\text{m}\text{o}\text{l}}=12.4\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\\ \text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{25.0\text{g}}{\text{28}\text{.013 }\text{g}/\text{m}\text{o}\text{l}}=0.892\text{m}\text{o}\text{l}\text{e}\text{s}{\text{N}}_2\end{array}$

In this reaction ${\text{N}}_2$ is a limiting reactant because it completely reacted in the reaction.

Amount of product in gram calculated as follows:

$0.892\text{m}\text{o}\text{l}\text{e}\text{s}{\text{N}}_2\times \frac{\cancel{2.00\text{m}\text{o}\text{l}\text{e}\text{N}{\text{H}}_3}}{1.00\text{m}\text{o}\text{l}\text{e}\text{s}{\text{N}}_2}\times \frac{\text{17}\text{.031 }\text{g}\text{N}{\text{H}}_3}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{N}{\text{H}}_3}}=30.38\text{g}\text{N}{\text{H}}_3$.