Question

Urea $\left(H_{2}NCON{H}_2\right)$ is manufactured by passing $C{O}_2\left(g\right)$ through ammonia solution followed by crystallization. $C{O}_2$ for the above reaction is prepared by combustion of hydrocarbon. If combustion of $236kg$ of a saturated hydrocarbon $\left(C_n{H}_{2n+2}\right)$ produces as much $C{O}_2$ as required for production of $999.6kg$ urea then molecular formula of hydrocarbon is:

• A. $C_{10}{H}_{22}$
• B. $C_{12}{H}_{26}$
• C. $C_{13}{H}_{28}$
• D. $C_8{H}_{18}$

Answer 1

The correct option is B $C_{12}{H}_{26}$

$2N{H}_3+C{O}_2\to N{H}_2-CO-N{H}_2+H_{2}O\text{moles of urea}=\frac{999.6\times {10}^3}{60}=16.66\times {10}^3\therefore \text{moles of}C{O}_2=16.66\times {10}^3{C}_n{H}_{2n+2}+O_2\to nC{O}_2+H_{2}O\frac{16.66\times {10}^3}{n}\times [12n+(2n+2\left)\right]=236\times {10}^{3}n\approx 12$
$\therefore \text{molecular formula of hydrocarbon}=C_{12}{H}_{26}$