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Question

A cylinder contains either ethylene or propylene $$12\ ml$$ of gas required $$54\ ml$$ of oxygen for complete combustion. The gas is ______.

A
Ethylene
B
Propylene
C
$$1:1$$ mixture of two gases
D
$$1:2$$ mixture
Solution
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Correct option is B. Propylene
$$\textbf{Step 1}$$ Chemical reaction for the combustion of ethylene and propylene,

$$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$$
$$C_3H_6+\frac {9}{2}O_2\rightarrow 3CO_2+3H_2O$$
1 mol of $$C_2H_4=28 g=22400\ mL$$
1 mole of $$C_3H_6=42 g=22400\ mL$$
1 mole of $$O_2=32g=22400\ mL$$

$$\textbf{Step 2}$$ Calculation of oxygen require in the combusion of ethylene
As 1 mol ($$22400\ mL$$) of ethylene require 3 mol ($$3\times 22400\ mL$$) of oxygen
$$\therefore 12\ mL$$ ethylene will need $$3\times 12$$ mL $$O_2$$ = $$36\ mL$$ of $$O_2$$

$$\textbf{Step 3}$$ Calculation of oxygen require in the combusion of propylene.
As 1 mol ($$22400\ mL$$) of propylene requires 3 mol ($$3\times 22400\ mL$$) of oxygen
$$\therefore 12\ mL$$ propylene will need $$4.5\times 12$$ mL $$O_2$$ = $$54\ mL$$ $$O_2$$
Therefore option B is correct.

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