Question

$20.0 \,kg$ of $N_2(g)$ and $3.0\, kg$ of $H_2(g)$ are mixed to produce $NH_3(g)$. The amount of $NH_3(g)$ formed is

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

Answer 1

Answer: (A)

$17\, kg$

Answer 2

$N _{2}+3 H _{2} {<=>} 2 NH _{3}$
$\frac{20 \times 10^{3} g}{28} \frac{3 \times 10^{3} g}{2}$
$=711.2$ mole. $1500$ mole
$L.R= \frac{711.2}{1} = \frac{1500}{3}$
$3 mole \longrightarrow 2 mole$
$1 mole \longrightarrow \frac{2}{5} mole$
$1500 mole \longrightarrow \frac{2}{3}mole$
$= 1000$ mole
$17 \times 1000$
$= 17\,kg$