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$56 L t$ of $N_{2}$ gas is mixed with excess of $H_{2}$ gas and is found that $20 L t$ of $N H_{3}$ gas is produced. The volume of unused $N_{2}$ gas is found to be__ Lt.

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$49.4 L t$

$46 L t$

$52 L t$

$28 L t$

answer is B.

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Detailed Solution

$N_{2} + 3 H_{2} \to 2 N H_{3}$

$56 L t -$

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