Question

A possible practical way to eliminate oxides of nitrogen from automobile exhaust gases uses Cyanuric acid (${\mathit{C}}_{\mathbf{3}}{\mathit{N}}_{\mathbf{3}}{\left(OH\right)}_{\mathbf{3}}$). When heated to the relatively low temperature of $\mathbf{62}\mathbf{.}\mathbf{5}\mathbf{}{}^{\mathbf{°}}\mathit{F}$, Cyanuric acid reacts with in the exhaust to form nitrogen, carbon dioxide and water all of which are normal constituents of the air.

a. Write balanced equation for these two reactions

b. If the process described earlier became practical, how much cyanide acid (in kilograms) would be required to absorb the $\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathit{K}\mathit{g}\mathbf{}\mathit{N}{\mathit{O}}_{\mathbf{2}}$generated annually in auto exhaust in the United States?

Short answer

a. Balanced equation for these two reactions are:

$$\begin{gathered} C_3{N}_3(OH{)}_3\to 3HCNO \\ 8HCNO+6N{O}_2\to 7N_2+8C{O}_2+4H_{2}O \end{gathered}$$

b. The mass of Cyanuric acid required to absorb $1.7\times {10}^{10}kgN{O}_2$is$2.1\times {10}^{10}kg$

Step-by-step solution

Cyanuric acid chemical reaction at 62.5F°

Cyanuric acid chemical reaction

$C_3{N}_3(OH{)}_3\to 3HCNO$

(Cyanuric acid) (Isocyanic acid)

Cyanuric acid at$62.5{}^{°}F$fromIsocyanic acid

Balanced Equation is

$C_3{N}_3{\left(OH\right)}_3\to 3HCNO$

Isocyanic acid reaction

Isocyanic acid reaction with $N{O}_2$

$HCNO+N{O}_2\to N_2+C{O}_2+H_{2}O$

Isocyanic acid react with$N{O}_2$ to form Nitrogen, Carbon Dioxide and Water

Balanced equation is

$8HCNO+N{O}_2\to N_2+8C{O}_2+4H_{2}O$

Now balancing$O_2$ atoms on both the sides

$8HCNO+6N{O}_2\to 7N_2+8C{O}_2+4H_{2}O$

Required mass of Cyanuric Acid

Molar mass of$C_3{N}_3{\left(OH\right)}_3$IS$129.08gmo{l}^{-1}$

Molar mass of$46.01gmo{l}^{-1}$IS$46.01gmo{l}^{-1}$

According to balanced reactions B moles of HCNO produced from one mole of $C_3{N}_3{\left(OH\right)}_3$and 8 moles of HCNO absorb 6 moles of $N{O}_2$

Number of moles of HCNO required for absorb $1.7\times {10}^{10}kgN{O}_2$is

$\left(1.7\times {10}^{10}kgN{O}_2\right)\times \frac{10g}{1kg}\times \frac{1molN{O}_2}{46.01gN{O}_2}=4.9\times {10}^{11}molHCNO$

Required mass of Cyanuric acid required to produce $4.1\times {10}^{11}molHCNO$is

$$\begin{gathered} 4.1\times {10}^{11}molHCNO\times \frac{1mol{C}_3{N}_3{\left(OH\right)}_3}{3molHCNO}\times \frac{129.08g{C}_3{N}_3{\left(OH\right)}_3}{1mol{C}_3{N}_3{\left(OH\right)}_3} \\ =2.1\times {10}^{13}g{C}_3{N}_3{\left(OH\right)}_3 \\ =2.1\times {10}^{10}g{C}_3{N}_3{\left(OH\right)}_3 \end{gathered}$$

Hence the mass of Cyanuric acid required is$2.1\times {10}^{10}$ kg