Question

When $0.273g$ of $Mg$ is heated strongly in a nitrogen $\left(N_2\right)$ atmosphere, $0.378g$ of the compound is formed. Hence, the compound formed is :

• A. $M{g}_3{N}_2$
• B. $M{g}_{3}N$
• C. $M{g}_2{N}_3$
• D. $MgN$

Answer 1

Answer: (A)

$M{g}_3{N}_2$

The number of moles of Mg are $\frac{0.273}{24.3}=0.0112mol$

The number of moles of nitrogen are $\frac{0.378-0.273}{14}=7.5\times {10}^{-3}mol$

The mole ratio is $\frac{0.0112}{7.5\times {10}^{-3}}=1.5:1=3:2$

Hence, the formula is $M{g}_3{N}_2$