The mass of Mg3N2 produced if 48 g of Mg metal is reacted with 34 g NH3 gas is Mg + NH3 → Mg3N2 + H2

Engineering

Chemistry

Chemical Equation and Stoichiometric Calculation

Question

The mass of Mg₃N₂ produced if 48 g of Mg metal is reacted with 34 g NH₃ gas is
Mg + NH₃ → Mg₃N₂ + H₂

$\frac{100}{3}$

$\frac{150}{3}$

$\frac{200}{3}$

$\frac{400}{3}$

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3Mg + 2NH₃ → Mg₃N₂ + 3H₂
mol $\frac{48}{24}$ = 2 $\frac{34}{17} = 2$
L.R. Mg

Mass of Mg₃N₂ = $\frac{1}{3}$ × 2 × (3 × 24 + 28)

$= \frac{200}{3}$