Question
The ratio of mass percentage $(w/w)$ of $C:H$ in a hydrocarbon is $12:1$. It has two carbon atoms. The weight (in g ) of ${CO}_{2}(\text{ }g)$ formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is :
(Given: Molar mass in $g{mol}^{- 1}C:12,H:1,O:16$ )
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Solution
Sol. [Key Concept: Mass ratio C:H $= 12:1$ with 2 carbons $\Rightarrow$ C$2$H$2$; moles $\rightarrow$ mass of CO$2$.]
Molecular formula of hydrocarbon is $C_{2}H_{2}$
Moles of $C_{2}H_{2} = \frac{3.38}{26}$
$$C_{2}H_{2}(\text{ }g) + \frac{5}{2}O_{2}(\text{ }g) \rightarrow 2{CO}_{2}(\text{ }g) + H_{2}O(l) $$Moles of ${CO}_{2}$ produced $= \left( \frac{3.38}{26} \times 2 \right)$
Mass of ${CO}_{2}$ produced $= \left( \frac{3.38}{26} \times 2 \right) \times 44 = 11.44$
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