Mole ConceptHard
Question
When a hydrocarbon is burnt completely, the ratio of masses of CO2 and H2O formed is 44: 27. The hydrocarbon is
Options
A.CH4
B.C2H6
C.C2H4
D.C2H2
Solution
$C_{x}H_{y} + \left( x + \frac{y}{4} \right)O_{2} \rightarrow xCO_{2} + \frac{y}{2}H_{2}O$
$44x\text{ g} \frac{y}{2} \times 18 = 9y\text{ g}$
From question, $\frac{44x}{9y} = \frac{44}{27} \Rightarrow \frac{x}{y} = \frac{1}{3}$
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