Urea $(H_{2} N C O N H_{2})$ is manufactured by passing $C O_{2} (g)$ through ammonia solution followed by crystallization. $C O_{2}$ for the above reaction is prepared by combustion of hydrocarbon. If combustion of $236 k g$ of a saturated hydrocarbon $(C_{n} H_{2 n + 2})$ produces as much $C O_{2}$ as required for production of $999.6 k g$ urea then molecular formula of hydrocarbon is:

C_{10} H_{22}

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C_{12} H_{26}

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C_{13} H_{28}

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C_{8} H_{18}

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Solution

The correct option is B $C_{12} H_{26}$
$2 N H_{3} + C O_{2} \to N H_{2} - C O - N H_{2} + H_{2} O moles of urea = \frac{999.6 \times 10^{3}}{60} = 16.66 \times 10^{3} ∴ moles of C O_{2} = 16.66 \times 10^{3} C_{n} H_{2 n + 2} + O_{2} \to n C O_{2} + H_{2} O \frac{16.66 \times 10^{3}}{n} \times [12 n + (2 n + 2)] = 236 \times 10^{3} n \approx 12$
$∴ molecular formula of hydrocarbon = C_{12} H_{26}$

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