One mole of a hydrocarbon, on combustion in air, produces two moles of $ \text{CO}_2 $, two moles of $ \text{H}_2\text{O} $, and a large amount of heat. This hydrocarbon is:

Question

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Let the hydrocarbon be represented by $C_xH_y$. The balanced combustion reaction is: $$ C_xH_y + \left( x + \frac{y}{4} \right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O $$ We are given that one mole of the hydrocarbon produces two moles of $CO_2$ and two moles of $H_2O$. Therefore, $x = 2$ and $\frac{y}{2} = 2$, which means $y = 4$. The hydrocarbon is $C_2H_4$, which is ethane.

Answer

<div>ethyne</div>

Answer

<div>ethene</div>

Answer

<div>ethane</div>

Answer

<div>methane</div>

One mole of a hydrocarbon, on combustion in air, produces two moles of \( \text{CO}_2 \), two moles of \( \text{H}_2\text{O} \), and a large amount of heat. This hydrocarbon is:

The Correct Option is C

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