Chapter 3, Problem 3.111QP

(a)

Interpretation Introduction

Interpretation: The theoretical yield of the products of each of the given unbalanced chemical equations for the given reactions is to be determined.

Concept introduction: The mass of product formed in the given reaction depends upon the mass of the limiting reagent in the reaction.

To determine: The mass of product formed in the given unbalanced reaction.

Answer to Problem 3.111QP

Solution

The mass of ${\text{Li}}_3\text{N}$ formed in the given reaction is $4.97g$.

Explanation of Solution

Explanation

Given

The unbalanced chemical equation for the given reaction is,

$\text{Li}\left(s\right)+{\text{N}}_2\left(g\right)\to {\text{Li}}_3\text{N}\left(s\right)$

The mass of lithium in the reaction is $5.0\text{g}$.

The mass of nitrogen gas in the reaction is $2.0\text{g}$.

The given reaction contains one lithium atom and two nitrogen atoms on the reactant side and three lithium atoms and one nitrogen atom on the product side. Therefore, the given reaction is balanced by multiplying the $\text{Li}$ with $6$ and ${\text{Li}}_3\text{N}$ with $2$.

$\text{6Li}\left(s\right)+{\text{N}}_2\left(g\right)\to 2{\text{Li}}_3\text{N}\left(s\right)$

The number of moles of the given element is calculated by using the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$ (1)

Substitute the values of mass and molar mass of lithium in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{Li}=\frac{5.00\text{g}}{6.94\text{g/mol}}\\ =0.7204\text{mol}\end{array}$

Substitute the values of mass and molar mass of nitrogen in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{N}}_2=\frac{2.00\text{g}}{28.014\text{g/mol}}\\ =0.0714\text{mol}\end{array}$

The number of moles of nitrogen required in the given reaction is,

$\begin{array}{c}\text{Moles}\text{of}{\text{N}}_2=\frac{1}{6}\times 0.7204\text{mol}\left(\text{Li}\right)\\ =0.120\text{mol}\end{array}$

That is more than the number of moles of nitrogen present in the reaction medium. Therefore, nitrogen is the limiting reagent in the given reaction.

Since, the number of moles of ${\text{N}}_2\left(g\right)$ is equal to the half of the moles of ${\text{Li}}_3\text{N}$ in the given reaction. Therefore, the number of moles of ${\text{Li}}_3\text{N}$ formed in the given reaction is double as compared to ${\text{N}}_2\left(g\right)$ as the reaction stops immediately after the concentration of ${\text{N}}_2\left(g\right)$ finishes.

The number of moles of ${\text{Li}}_3\text{N}$ formed is $2\times 0.0714\text{mol}=0.1428\text{mol}$.

The mass of ${\text{Li}}_3\text{N}$ formed in the given reaction is calculated by multiplying its number of moles with the molar mass.

$\begin{array}{c}\text{Mass}\text{of}{\text{Li}}_3\text{N}=0.1428\text{mol}\times 34.827\text{g/mol}\\ =4.97\text{g}\end{array}$

Thus, the mass of ${\text{Li}}_3\text{N}$ formed in the given reaction is $4.97g$.

(b)

Interpretation Introduction

To determine: The mass of product formed in the given unbalanced reaction.

Answer to Problem 3.111QP

Solution

The mass of ${\text{H}}_3{\text{PO}}_4$ formed in the given reaction is $34.5g$.

Explanation of Solution

Explanation

Given

The unbalanced chemical equation for the given reaction is,

${\text{P}}_2{\text{O}}_5\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{PO}}_4\left(aq\right)$

The mass of ${\text{P}}_2{\text{O}}_5$ in the reaction is $25.0\text{g}$.

The mass of water in the reaction is $36.0\text{g}$.

The given reaction contains two phosphorus atoms, six oxygen atoms and two hydrogen atoms on the reactant side and three hydrogen atoms, one phosphorus atom and four oxygen atoms on the product side. Therefore, the given reaction is balanced by multiplying the ${\text{H}}_2\text{O}$ with $3$ and ${\text{H}}_3{\text{PO}}_4$ with $2$.

${\text{P}}_2{\text{O}}_5\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)\to 2{\text{H}}_3{\text{PO}}_4\left(aq\right)$

The number of moles of the given element is calculated by using the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$ (1)

Substitute the values of mass and molar mass of ${\text{P}}_2{\text{O}}_5$ in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{P}}_2{\text{O}}_5=\frac{25.0\text{g}}{141.941\text{g/mol}}\\ =0.1761\text{mol}\end{array}$

Substitute the values of mass and molar mass of ${\text{H}}_2\text{O}$ in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{H}}_2\text{O}=\frac{36.0\text{g}}{18.02\text{g/mol}}\\ =1.998\text{mol}\end{array}$

The number of moles of ${\text{P}}_2{\text{O}}_5$ required in the given reaction is,

$\begin{array}{c}\text{Moles}\text{of}{\text{P}}_2{\text{O}}_5=\frac{1}{3}\times 1.998\text{mol}\left({\text{H}}_2\text{O}\right)\\ =0.666\text{mol}\end{array}$

That is more than the number of moles of ${\text{P}}_2{\text{O}}_5$ present in the reaction medium. Therefore, ${\text{P}}_2{\text{O}}_5$ is the limiting reagent in the given reaction.

Since, the number of moles of ${\text{P}}_2{\text{O}}_5$ is equal to the half of the moles of ${\text{H}}_3{\text{PO}}_4$ in the given reaction. Therefore, the number of moles of ${\text{H}}_3{\text{PO}}_4$ formed in the given reaction is double as compared to ${\text{P}}_2{\text{O}}_5$ as the reaction stops immediately after the concentration of ${\text{P}}_2{\text{O}}_5$ finishes.

The number of moles of ${\text{H}}_3{\text{PO}}_4$ formed is $2\times 0.1761\text{mol}=0.3522\text{mol}$.

The mass of ${\text{H}}_3{\text{PO}}_4$ formed in the given reaction is calculated by multiplying its number of moles with the molar mass.

$\begin{array}{c}\text{Mass}\text{of}{\text{H}}_3{\text{PO}}_4=0.3522\text{mol}\times 97.993\text{g/mol}\\ =34.5\text{g}\end{array}$

Thus, the mass of ${\text{H}}_3{\text{PO}}_4$ formed in the given reaction is $34.5g$.

(c)

Interpretation Introduction

To determine: The mass of product formed in the given unbalanced reaction.

Answer to Problem 3.111QP

Solution

The mass of ${\text{SO}}_3$ formed in the given reaction is $7.93g$.

Explanation of Solution

Explanation

Given

The unbalanced chemical equation for the given reaction is,

${\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{SO}}_3\left(g\right)$

The mass of ${\text{SO}}_2$ in the reaction is $6.4\text{g}$.

The mass of ${\text{O}}_2$ in the reaction is $4.0\text{g}$.

The given reaction contains one sulfur atom and four oxygen atoms on the reactant side and one sulfur atom and three oxygen atoms on the product side. Therefore, the given reaction is balanced by multiplying the ${\text{SO}}_2$ with $2$ and ${\text{SO}}_3$ with $2$.

${\text{2SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{SO}}_3\left(g\right)$

The number of moles of the given element is calculated by using the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$ (1)

Substitute the values of mass and molar mass of ${\text{SO}}_2$ in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{SO}}_2=\frac{6.4\text{g}}{64.058\text{g/mol}}\\ =0.099\text{mol}\end{array}$

Substitute the values of mass and molar mass of ${\text{O}}_2$ in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{O}}_2=\frac{4.0\text{g}}{31.998\text{g/mol}}\\ =0.125\text{mol}\end{array}$

The number of moles of ${\text{SO}}_2$ required in the given reaction is,

$\begin{array}{c}\text{Moles}\text{of}{\text{SO}}_2=2\times 0.125\text{mol}\left({\text{O}}_2\right)\\ =0.25\text{mol}\end{array}$

That is more than the number of moles of ${\text{SO}}_2$ present in the reaction medium. Therefore, ${\text{SO}}_2$ is the limiting reagent in the given reaction.

Since, the number of moles of ${\text{SO}}_2$ is equal to the number of moles of ${\text{SO}}_3$ in the given reaction. Therefore, the number of moles of ${\text{SO}}_2$ formed in the given reaction is equal to that of ${\text{SO}}_2$ as the reaction stops immediately after the concentration of ${\text{SO}}_2$ finishes.

The number of moles of ${\text{SO}}_3$ formed is $0.099\text{mol}$.

The mass of ${\text{SO}}_3$ formed in the given reaction is calculated by multiplying its number of moles with the molar mass.

$\begin{array}{c}\text{Mass}\text{of}{\text{SO}}_3=0.099\text{mol}\times 80.057\text{g/mol}\\ =7.93\text{g}\end{array}$

Thus, the mass of ${\text{SO}}_3$ formed in the given reaction is $7.93g$.

Conclusion

1. a. The mass of ${\text{Li}}_3\text{N}$ formed in the given reaction is $4.97g$.
2. b. The mass of ${\text{H}}_3{\text{PO}}_4$ formed in the given reaction is $34.5g$.
3. c. The mass of ${\text{SO}}_3$ formed in the given reaction is $7.93g$