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, the molecular formula of the hydrocarbon is :

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

Answer 1

Answer: (B)

$C_{12}{H}_{26}$

$2N{H}_3+C{O}_2\to N{H}_{2}CON{H}_2+H_{2}O$
Moles of urea$=\frac{999.6\times {10}^3}{60}=16.66\times {10}^3$.
$\therefore$ Moles of $C{O}_2=16.66\times {10}^3$.
$C_n{H}_{2n+2}+O_2\to nC{O}_2+H_{2}O$
$\frac{16.66\times {10}^3}{n}\times [12n+(2n+2\left)\right]=236\times {10}^3$
$n\cong 12$

$\therefore$ Molecular formula of the hydrocarbon$=C_{12}{H}_{26}$.