Chapter 3, Problem 140AP

Interpretation Introduction

Interpretation:

The mass of ammonia is to be calculated with given mass of nitric acid.

Concept introduction:

The number of moles is defined as the ratio of mass to the molar mass:

$n=\frac{m}{M}$

Here, $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.

Molar mass is calculated by adding the masses of each element multiplied by the number of atoms present (given in subscript). Its S.I. unit is $\text{g}/\text{mol}$.

The mass of compound can be calculated as: $m=n\times M$

Here, $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.

The relationship between pounds and grams can be expressed as $\text{1 lb}\text{=}\text{453}\text{.6 g}$.

To convert pounds to grams, conversion factor is $\frac{453.6\text{ g}}{\text{1 lb}}$.

Answer to Problem 140AP

Solution: $9.58\times {10}^5\text{ g}$

Explanation of Solution

Given information:

$m_{{\text{HNO}}_3}=1\text{ ton}$

The formation of ${\text{HNO}}_3$ takes place by the Ostwald process with the help of three reactions as follows:

The reaction for formation of $\text{NO}$ is

$4{\text{ NH}}_3\left(g\right)+5{\text{ O}}_2\left(g\right)\to 4\text{ NO}\left(g\right)+6\text{}{\text{ H}}_{\text{2}}\text{O}\left(g\right)$

The reaction for formation of ${\text{NO}}_2$ is

$2\text{ NO}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{ NO}}_2\left(g\right)$

The reaction for formation of ${\text{HNO}}_3$ is

$2{\text{ NO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{HNO}}_3\left(aq\right)+{\text{HNO}}_2\left(aq\right)$

The mass of ${\text{HNO}}_3$ is $1\text{ ton}$.

In one ton, there are $\text{2000 lb}$.

In one pound, there are $453.6\text{ g}$.

Convert pound to gram:

$\begin{array}{c}\text{2000 lb}=\left(\text{2000 }\cancel{\text{lb}}\right)\left(\frac{453.6\text{ g}}{\text{1 }\cancel{\text{lb}}}\right)\\ =907200\text{ g}\end{array}$

So, the mass of ${\text{HNO}}_3$ is $9.072\times {10}^5\text{ g}$

The molar mass of ${\text{HNO}}_3$ is $63\text{ g}/\text{mol}$

Calculate the number of moles of ${\text{HNO}}_3$ as

$n_{{\text{HNO}}_3}=\frac{m_{{\text{HNO}}_3}}{M_{{\text{HNO}}_3}}$

Substitute $9.072\times {10}^5\text{ g}$ for $m$ and $63\text{ g}/\text{mol}$ for $M$ in the equation above

$\begin{array}{c}{n}_{{\text{HNO}}_3}=\frac{9.072\times {10}^5\text{ g}}{63\text{ g}/\text{mol}}\\ =1.44\times {10}^4\text{ mol}\end{array}$

Now, in the reaction for formation of ${\text{HNO}}_3$, the mole ratio of ${\text{NO}}_2$ and ${\text{HNO}}_3$ is $2:1$ and the yield is only $80\%$.

So, the number of moles of ${\text{NO}}_2$ is

$n_{{\text{NO}}_2}=\frac{2\times n_{{\text{HNO}}_3}}{0.8}$

Substitute $1.44\times {10}^4\text{ mol}$ for $n_{{\text{HNO}}_3}$ in the equation above

$\begin{array}{c}{n}_{{\text{NO}}_2}=2\times \frac{1.44\times {10}^4\text{ mol}}{0.8}\\ =\frac{2.88\times {10}^4\text{ mol}}{0.8}\\ =3.6\times {10}^4\text{ mol}\end{array}$

In the reaction for formation of ${\text{NO}}_2$, the mole ratio of ${\text{NO}}_2$ and $\text{NO}$ is $2:2$ and the yield is only $80\%$.

So, the number of moles of $\text{NO}$ is

$n_{\text{NO}}=\frac{n_{{\text{NO}}_2}}{0.8}$

Substitute $3.6\times {10}^4\text{ mol}$ for $n_{{\text{NO}}_2}$ in the equation above

$\begin{array}{c}{n}_{\text{NO}}=\frac{3.6\times {10}^4\text{ mol}}{0.8}\\ =4.5\times {10}^4\text{ mol}\end{array}$

Similarly, in the reaction for formation of $\text{NO}$, the mole ratio of $\text{NO}$ and ${\text{NH}}_3$ is $4:4$ and the yield is only $80\%$.

So, the number of moles of ${\text{NH}}_3$ is

$n_{{\text{NH}}_3}=\frac{n_{\text{NO}}}{0.8}$

Substitute $4.5\times {10}^4\text{ mol}$ for $n_{\text{NO}}$ in the equation above

$\begin{array}{c}{n}_{{\text{NH}}_3}=\frac{4.5\times {10}^4\text{ mol}}{0.8}\\ =5.62\times {10}^4\text{ mol}\end{array}$

The molar mass of ammonia is $17\text{ g}/\text{mol}$.

So, the mass of ${\text{NH}}_3$ required to produce $1\text{ ton}$ of ${\text{HNO}}_3$ is

$m_{{\text{NH}}_3}=n_{{\text{NH}}_3}\times M_{{\text{NH}}_3}$

Substitute $5.62\times {10}^4\text{ mol}$ for $n_{{\text{NH}}_3}$ and $17\text{ g}/\text{mol}$ for $M_{{\text{NH}}_3}$ in the equation above

$\begin{array}{c}{m}_{{\text{NH}}_3}=5.62\times {10}^4\text{ mol}\times 17\text{ g}/\text{mol}\\ =95.8\times {10}^4\text{ g}\\ =9.58\times {10}^5\text{ g}\end{array}$

Conclusion

The mass of ammonia required is $9.58\times {10}^5\text{ g}$.