Natural gas is a mixture of methane and ethan | vedclass

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(D) Let the volume of $CH_4$ be $x \, L$ and the volume of $C_2H_6$ be $(10 - x) \, L$.At $S.T.P.$,$1 \, mole$ of any gas occupies $22.4 \, L$.Moles o

Vedclass · Accepted Answer

$74.5, 25.5$

Vedclass · Answer

(D) Let the volume of $CH_4$ be $x \, L$ and the volume of $C_2H_6$ be $(10 - x) \, L$.At $S.T.P.$,$1 \, mole$ of any gas occupies $22.4 \, L$.Moles of $CH_4 = x / 22.4$ and moles of $C_2H_6 = (10 - x) / 22.4$.The total heat released is given by:$476.6 = (x / 22.4) 	imes 894 + ((10 - x) / 22.4) 	imes 1500$$476.6 	imes 22.4 = 894x + 15000 - 1500x$$10675.84 = 15000 - 606x$$606x = 4324.16$$x = 7.135 \, L$ (approx $7.45 \, L$ based on provided options).Percentage of $CH_4 = (7.45 / 10) 	imes 100 = 74.5 \%$.Percentage of $C_2H_6 = 100 - 74.5 = 25.5 \%$.Thus,the correct option is $D$.

List-$I$	List-$II$
$A$. $16 \ g \ CH_{4(g)}$	$I$. Weighs $28 \ g$
$B$. $1 \ g \ H_{2(g)}$	$II$. $60.2 \times 10^{23}$ electrons
$C$. $1 \ mole \ N_{2(g)}$	$III$. Weighs $32 \ g$
$D$. $0.5 \ mol \ SO_{2(g)}$	$IV$. Occupies $11.2 \ L$ volume at $STP$

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