Question

Nitric acid is produced commercially by the Ostwald process. The three steps of the Ostwaldprocess are shown in the following equations:
${\mathbf{\text{4NH}}}_{\mathbf{3}}\left(g\right)\mathbf{+}\mathbf{5}{\mathbf{\text{O}}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{4}\mathbf{\text{NO}}\left(g\right)\mathbf{+}\mathbf{6}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\left(g\right)$

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

$\mathbf{3}{\mathbf{\text{NO}}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\left(l\right)\mathbf{\to }\mathbf{2}{\mathbf{\text{HNO}}}_{\mathbf{3}}\left(aq\right)\mathbf{+}\mathbf{\text{NO}}\left(g\right)$

What mass of${\mathbf{\text{NH}}}_{\mathbf{3}}$must be used to produce$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{kg}}$of${\mathbf{\text{HNO}}}_{\mathbf{3}}$by the Ostwald process, assumingyield in each reaction and assuming the$\mathbf{\text{NO}}$produced in the third stage is notrecycled?

Short answer

The mass of${\text{NH}}_3$ must be used to produce $1.0\times {10}^6\mathrm{kg}$of ${\text{HNO}}_3$by the Ostwald process is$\mathbf{0}\mathbf{.}\mathbf{4029}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{kg}}$ .

Step-by-step solution

Definition

The mass of any materialexpressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of number of moles

Molar mass of $HN{O}_3=\left(1.0079\right)+\left(14.00\right)+3\left(16.00\right)$

$=63.00\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

$\mathbf{\text{Numberofmoles}}\mathbf{=}\frac{\mathbf{\text{Mass}}}{\mathbf{\text{Molarmass}}}$

$=1.0\times {10}^6\times \frac{{10}^3}{63.00}$

$=0.0158\times {10}^9\mathrm{Moles}$

Finding the number of moles of  NO2 needed for HNO3

From the given reaction,

$3{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)\to 2{\mathrm{HNO}}_3\left(\mathrm{aq}\right)+\mathrm{NO}\left(\mathrm{g}\right)$

So, for 2 moles of ${\text{HNO}}_3$, 3 moles of ${\text{NO}}_2$required

Now, for$0.0158\times {10}^9$ moles of ${\text{HNO}}_3$

$=\frac{0.0158\times {10}^9}{2}\times 3$

$=0.0237\times {10}^9\mathrm{Moles}$ of ${\text{NO}}_2$needed

Finding the number of NOmoles of needed for  0.0237×109Moles of NO2

From the given reaction,

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

So, for 2 moles of ${\mathbf{\text{NO}}}_{\mathbf{2}}$, 3 moles of $\mathbf{\text{NO}}$required

Now, for $\mathbf{0}\mathbf{.}\mathbf{0237}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$moles of ${\mathbf{\text{NO}}}_{\mathbf{2}}$, $\mathbf{0}\mathbf{.}\mathbf{0237}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$moles of $\mathbf{\text{NO}}$ required.

Finding the number of moles of NH3 needed for 0.0237×109 Moles of NO

From the given reaction,

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

So, for 4 molesof $\mathbf{\text{NO}}$, 4 moles of ${\mathbf{\text{NH}}}_{\mathbf{3}}$required

Now, for $\mathbf{0}\mathbf{.}\mathbf{0237}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$moles of $\mathbf{\text{NO}}$, $\mathbf{0}\mathbf{.}\mathbf{0237}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$moles of required${\mathbf{\text{NH}}}_{\mathbf{3}}$

Determining the mass of Ammonia

$\mathbf{\text{MassofAmmonia}}\mathbf{=}\left(\text{Mass}\right)\mathbf{\times }\left(\text{Molarmass}\right)$

$=\left(0.0237\times {10}^9\right)\left(17\right)$

$=0.4029\times {10}^{9}g$

Therefore, the mass of ammonia produced is$\mathbf{0}\mathbf{.}\mathbf{4029}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{kg}}$