Question

Bacterial digestion is an economical method of sewage treatment. Thereaction

$$\begin{gathered} \mathbf{5}\mathit{C}{\mathit{O}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{55}\mathit{N}{{\mathbf{H}}_{\mathbf{4}}}^{\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{76}{\mathit{O}}_{\mathbf{2}}\left(g\right){\mathit{C}}_{\mathbf{5}}{\mathit{H}}_{\mathbf{7}}{\mathit{O}}_{\mathbf{2}}\mathit{N}\left(s\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{54}\mathit{N}{{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{-}}\left(aq\right)\mathbf{+}\mathbf{52}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)\mathbf{+}\mathbf{109}{\mathit{H}}^{\mathbf{+}}\left(aq\right) \end{gathered}$$
is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions.How much bacterial tissue is produced in a treatment plant for every$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{k}\mathit{g}$of wastewatercontaining$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\%}\mathit{N}{\mathit{H}}_{\mathbf{4}}$ions by mass? Assume that$\mathbf{95}\mathbf{\%}$ of the ammonium ions are consumed bythe bacteria.

Short answer

The bacterial tissue produced in a treatment plant for every$1.0\mathrm{x}{10}^4\mathrm{kg}$ of wastewater containing$3.0\%{\mathrm{NH}}_4$ ions by mass is$\mathbf{3}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathit{g}$ .

Step-by-step solution

Definition

The mass of any material expressed in atomic mass unitsisquantitatively equivalent to its molar mass in grams per mole.

Calculate the mass of NH4 in wastewater

Waste water contains $3.0\%{\mathrm{NH}}_4$

${\mathbf{\text{MassofNH}}}_{\mathbf{4}}\mathbf{=}\mathbf{3}\mathbf{\%}\mathbf{\times }\mathbf{}\mathbf{\text{Massofwastewater}}$

$=3.0\%\times 1\times {10}^{4}kg$

$=\frac{3}{100}\times 1\times {10}^{4}kg$

$=300kg$

Thus, only was$\mathbf{95}\mathbf{\%}$consumed by bacteria.

Mass of ${\text{NH}}_4$consumed by bacteria $=\frac{95}{100}\times 300kg$

$=285\times {10}^{3}g$

Balanced Equation

The balanced equation is:

$5{\mathrm{CO}}_2\left(\mathrm{g}\right)+55\mathrm{N}{{\mathrm{H}}_4}^{+}\left(\mathrm{aq}\right)+76{\mathrm{O}}_2\left(g\right)\to C_5{H}_7{O}_{2}N\left(s\right)+5{\mathrm{CO}}_2\left(\mathrm{g}\right)+55\mathrm{N}{{\mathrm{H}}_4}^{+}\left(\mathrm{aq}\right)+76{\mathrm{O}}_2\left(g\right)\to C_5{H}_7{O}_{2}N\left(s\right)+54N{O}_2\left(aq\right)+52H_{2}O\left(l\right)+109H\left(aq\right)54N{O}_2\left(aq\right)+52H_{2}O\left(l\right)+109H\left(aq\right)$

Let’s convert ${\mathbf{\text{NH}}}_{\mathbf{4}}$consumed by bacteria to the number of moles:

${\mathrm{NH}}_4=285\times {10}^{3}gN{H_4}^{+}\frac{1mol}{18gN{H}_4}$

$=15.833\times {10}^{3}molN{H}_4$

Thus for the balance reactions,

Mol ratio is $\frac{1{\mathrm{molC}}_5{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{N}\left(\mathrm{s}\right)}{55{\mathrm{molNH}}_4}$

${\mathrm{C}}_5{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{N}=15.833\times {10}^3\mathrm{mol}{\mathrm{NH}}_4\times \frac{1{\mathrm{molC}}_5{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{N}}{55\mathrm{mol}{\mathrm{NH}}_4}$

$=0.288\times {10}^3\mathrm{mol}{\mathrm{C}}_5{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{N}$

Determination of molar mass

Now, the molar mass $\left(C_5{H}_7{O}_{2}N\right)=5\times 12+7\times 1+2\times 16+14$

$=114\frac{g}{mol}$

$\mathrm{m}\left({\mathrm{C}}_5{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{N}\right)=0.288\times {10}^3{\mathrm{molC}}_5{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{N}\frac{113\mathrm{g}}{1{\mathrm{molC}}_5{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{N}}$

$=3.25\times {10}^{4}g$

Therefore, overall $\mathbf{3}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{g}$bacterial tissue is produced.