Calculate the volume of ammonia gas produced at STP when 140 g of nitrogen gas reacts with 30 g hydrogen gas.
(Atomic mass: $N = 14 u, H = 1 u$ )

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Solution

$N_{2} + 3 H_{2} ⟶ 2 N H_{3}$
Number of moles of $N_{2}$ in $140 g = \frac{140 g}{28} = 5$ moles of $N_{2}$
Number of moles of $H_{2}$ in $30 g = \frac{30}{2} = 15$ moles of $H_{2}$
moles $N_{2} + 3 H_{2} ⟶ 2 N H_{3}$
at $t = 0$ $5$ $15$ -
after reaction $5 - 5 \times 1$ $15 - 3 \times 5$ $2 \times 5$
Thus $10$ moles of $N H_{3}$ produced after reaction
as number of moles= $\frac{volume of product}{volume at STP}$
$\Rightarrow 10 = \frac{V}{22.4 L} \Rightarrow V = 224 L$
Thus $224 L$ of ammonia is produced.

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Calculate the volume of ammonia gas produced at STP when 140 g of nitrogen gas reacts with 30 g hydrogen gas. (Atomic mass: N=14 u, H=1 u)

Calculate the volume of ammonia gas produced at STP when 140 g of nitrogen gas reacts with 30 g hydrogen gas.
(Atomic mass: $N = 14 u, H = 1 u$ )