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Q.

The mass of Mg₃N₂ produced if 48 g of Mg metal is reacted with 34 g NH₃ gas is:
$3 Mg + 2 {NH}_{3} ⟶ {Mg}_{3} N_{2} + 3 H_{2}$

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a

$\frac{200}{3}$

b

$\frac{100}{3}$

c

$\frac{150}{3}$

d

$\frac{400}{3}$

answer is A.

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Detailed Solution

$3 Mg + 2 {NH}_{3} \to {Mg}_{3} N_{2} + 3 H_{2}$

$Mole \frac{48}{24} = 2 \frac{34}{17} = 2$

$∴ Mass of \frac{1}{3} \times 2 \times (3 \times 24 + 28) = \frac{200}{3}$

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The mass of Mg3N2 produced if 48 g of Mg metal is reacted with 34 g NH3 gas is:3Mg+2NH3⟶Mg3N2+3H2