Question

$20.0kg$ of $N_{2}\left(g\right)$ and $3.0kg$ of $H_{2}\left(g\right)$ are mixed to produce $\left(NH\right)_{3}\left(\right.g\left.\right)$ . The amount of $\left(NH\right)_{3}\left(g\right)$ formed is :

• A. $17kg$
• B. $34kg$
• C. $20kg$
• D. $3kg$

Answer 1

Answer: (A)

$\underset{28 g}{N_{2}}+\underset{6 g}{3 H_{2}} \rightarrow \underset{34 g}{2 NH_{3}}$
Limiting reactant is $H_{2}$ .
Weight of $NH_{3}$ obtained from $6\times 10^{- 3}kg$ of $H_{2}$
$=34\times 10^{- 3}kg$
$\therefore $ Weight of $NH_{3}$ obtained from $3kg$ of $H_{2}$
$=\frac{34 \times 10^{- 3}}{6 \times 10^{- 3}}\times 3=17kgNH_{3}$