$N H_{3}$ is formed in the following steps,
$I$ : $C a + 2 C ⟶ C a C_{2}$ ( $50 %$ yield)
$I I$ : $C a C_{2} + N_{2} ⟶ C a C N_{2} + C$ ( $100 %$ yield)
$I I I$ : $C a C N_{2} + 3 H_{2} O ⟶ 2 N H_{3} + C a C O_{3}$ ( $50 %$ yield)
To obtain $2$ moles of $N H_{3}$ , calcium required is :

1

mol

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2

mol

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3

mol

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4

mol

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Solution

The correct option is D $4$ mol
$I$ : $C a + 2 C ⟶ C a C_{2}$ $50 %$ yield
$I I$ : $C a C_{2} + N_{2} ⟶ C a C N_{2} + C$ $100 %$ yield
$I I I$ : $C a C N_{2} + 3 H_{2} O ⟶ 2 N H_{3} + C a C O_{3}$ $50 %$ yield
According to the balanced reaction,
$2$ moles of $N H_{3} = 2$ moles of $C a C N_{2} = 2$ moles $C a C_{2} = 4$ moles of $C a$ .
$50 %$ yield $100 %$ yield $50 %$ yield.

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Similar questions

N H_{3}

C a + 2 C \to C a C_{2}

C a C_{2} + N_{2} \to C a C N_{2} + C

C a C N_{2} + 3 H_{2} O \to 2 N H_{3} + C a C O_{3}

N H_{3}

3 M \to 2 B + C

(yield = 70 %) (i)

5 M \to 5 D + 2 C (ii)

3 E + 4 F \to 2 G + 3 H G + 3 I \to 4 J + 7 K 4 J + 3 L \to 9 N + 2 O

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