20.0 kg of N2(g) and 3.0 kg of H2(g) are mixed to produce NH3(g). The amount of NH3(g) formed is

Question

20.0 kg of N2(g) and 3.0 kg of H2(g) are mixed to produce NH3(g). The amount of NH3(g) formed is (A) 17 kg (B) 34 kg (C) 20 kg (D) 3 kg

Tardigrade Admin · Accepted Answer

N2+3 H2 leftharpoons 2 N3 (20 × 103 g/28) (3 × 103 g/2) =711.2 mole. 1500 mole L. R=(711.2/1)=(1500/3) 3 mole longrightarrow 2 mole 1 mole longrightarrow (2/5) mole 1500 mole longrightarrow (2/3) mole =1000 mole 17 × 1000 =17 kg

Q. $20.0 k g$ of $N_{2} (g)$ and $3.0 k g$ of $H_{2} (g)$ are mixed to produce $N H_{3} (g)$ . The amount of $N H_{3} (g)$ formed is

$17 k g$

$34 k g$

$20 k g$

$3 k g$

$23 k g$

$N_{2} + 3 H_{2} ⇌ 2 N_{3}$
$\frac{20 \times 1 0 ^{3} g}{28} \frac{3 \times 1 0 ^{3} g}{2}$
$= 711.2$ mole. $1500$ mole
L. $R = \frac{711.2}{1} = \frac{1500}{3}$
3 mole $⟶ 2$ mole
1 mole $⟶ \frac{2}{5}$ mole
1500 mole $⟶ \frac{2}{3}$ mole
$= 1000$ mole
$17 \times 1000$
$= 17 k g$

Q. 20.0kg of N2​(g) and 3.0kg of H2​(g) are mixed to produce NH3​(g). The amount of NH3​(g) formed is

17kg

34kg

20kg

3kg

23kg

N2​+3H2​⇌2N3​ 2820×103g​23×103g​ =711.2 mole. 1500 mole L. R=1711.2​=31500​ 3 mole ⟶2 mole 1 mole ⟶52​ mole 1500 mole ⟶32​ mole =1000 mole 17×1000 =17kg