$20.0 \mathrm{kg}$ of ${\mathrm{N}}_2\left(\mathrm{g}\right)$ and $3.0 \mathrm{kg}$ of ${\mathrm{H}}_2\left(\mathrm{g}\right),$ are mixed to produce ${\mathrm{NH}}_3\left(\mathrm{g}\right).$ The amount of ${\mathrm{NH}}_3 \left(\mathrm{g}\right)$ formed is:

Solution:

Mol. wt. of $\mathrm{N}2 = 2 \mathrm{x} 14 = 28 \mathrm{g} {\mathrm{mol}}^{-}$ : mol. wt. of $\mathrm{H}2 = 2 \mathrm{x} 1 = 2\mathrm{g} \mathrm{mol}$ mol. wt. of ${\mathrm{NH}}_3 = 14 + (3 \mathrm{x} 1) = 17$ $\mathrm{g} {\mathrm{mol}}^{-}.$

The given equation is:

     $$\begin{gathered} {\mathrm{N}}_2 + 3{\mathrm{H}}_2 ⟶ 2{\mathrm{NH}}_3 ...\left(1\right) \\ 28 \mathrm{kg} 3\mathrm{x}2=6 \mathrm{kg} 2\mathrm{x}17=34 \mathrm{kg} \\ \mathrm{or} 14 \mathrm{kg} 3 \mathrm{kg} 17 \mathrm{kg} \end{gathered}$$

Since the data shown in equation $\left(1\right)$ pertains to $3.0 \mathrm{kg}$ of ${\mathrm{H}}_2$ and in the question also, $3.0 \mathrm{kg}$ of ${\mathrm{H}}_2$ reacts to form ${\mathrm{NH}}_3,$ so ${\mathrm{H}}_2$ is a limiting reactant. So, amount of ${\mathrm{NH}}_3$ formed will be $17 \mathrm{kg}.$ 

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