Question

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane:

$\mathbf{2}{\mathbf{\text{NH}}}_{\mathbf{3}}\left(g\right)\mathbf{+}\mathbf{3}{\mathbf{\text{O}}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{\text{CH}}}_{\mathbf{4}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{\text{HCN}}\left(g\right)\mathbf{+}\mathbf{6}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\left(g\right)$

If $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\text{kg}}$each of ${\mathbf{\text{NH}}}_{\mathbf{3}\mathbf{.}}{\mathbf{\text{O}}}_{\mathbf{2}}$, and ${\mathbf{\text{CH}}}_{\mathbf{4}}$are reacted, what mass of $\mathbf{\text{HCN}}$and of ${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}$will be produced, assuming 100% yield?

Short answer

The masses of $\text{HCN}$and${\text{H}}_2\text{O}$ of that will be produced are$\mathbf{2}\mathbf{.}\mathbf{81}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{g}}$ and$\mathbf{5}\mathbf{.}\mathbf{63}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{g}}$ .

Step-by-step solution

Definition

The massof any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of moles of NH3,O2and CH4

Let us consider the given reaction:

$2{\mathrm{NH}}_3\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)+2{\mathrm{CH}}_4\left(\mathrm{g}\right)\to 2\mathrm{HCN}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

If the reactant has $5.00\times {10}^3\text{kg}$amount to produce$\text{HCN}$and of${\text{H}}_2\text{O}$.

Moles of ammonia $=5.00\times {10}^3\mathrm{kg}{\mathrm{NH}}_3\times \frac{{10}^3\mathrm{g}}{1\mathrm{kg}}\times \frac{1\mathrm{mol}{\mathrm{NH}}_3}{17.03\mathrm{g}{\mathrm{NH}}_3}$

$=2.936\times {10}^5\mathrm{mol}{\mathrm{NH}}_3$

Moles of oxygen role="math" localid="1648715405063">=5.00×103kgO2×103g1kg×1molO232.00gO2

$=1.5625\times {10}^5\mathrm{mol}{\mathrm{O}}_2$

Moles of methane $=5.00\times {10}^3\mathrm{kg}{\mathrm{CH}}_4\times \frac{{10}^3\mathrm{g}}{1\mathrm{kg}}\times \frac{1\mathrm{mol}{\mathrm{CH}}_4}{16.04\mathrm{g}{\mathrm{CH}}_4}$

$=3.117\times {10}^5\mathrm{mol}{\mathrm{CH}}_4$

Determination of mass of  HCN 

From the given reaction, 3 moles of ${\text{O}}_2$reacts with 2 moles of${\text{NH}}_3$ ,2 moles of role="math" ${\text{CH}}_4$

$$\begin{gathered} =1.5625\times {10}^6{\text{molO}}_2\text{×}\frac{2{\text{molNH}}_4}{3{\text{molO}}_2} \\ {\text{=1.0417×10}}^5{\text{molNH}}_4 \end{gathered}$$

From the given reaction,2 moles of ${\text{NH}}_3$reacts with $\text{HCN}$

role="math" localid="1648714956134" $$\begin{gathered} {\text{MassofHCN=1.047×10}}^5\text{molHCN×}\frac{27.018\text{gHCN}}{1\text{molHCN}} \\ {\text{=1.0417×10}}^5{\text{molNH}}_4 \\ \text{≈}2.81\times {10}^6\text{gHCN} \end{gathered}$$

From the given reaction,2 moles of ${\text{NH}}_3$reacts with ${\text{H}}_2\text{O}$

${\text{1.0417×10}}^5{\text{molNH}}_3\text{×}\frac{6{\text{molNH}}_2\text{O}}{2{\text{molNH}}_3}{\text{=1.0417×10}}^5{\text{molNH}}_3$

Mass of ${\text{H}}_2{\text{O=3.125×10}}^5{\text{molH}}_2\text{O×}\frac{18.02{\text{gH}}_2\text{O}}{1{\text{molH}}_2\text{O}}$

$=5.63\times {10}^6{\text{gH}}_2\text{O}$

Therefore,the masses of $\text{HCN}$ and${\text{H}}_2\text{O}$ are$\mathbf{2}\mathbf{.}\mathbf{81}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{\text{g}}$ and ${\mathbf{\text{5.63×10}}}^{\mathbf{6}}\mathbf{\text{g}}$.