Lithium metal reacts with nitrogen gas to produce a white solid Lithium nitride ( ${Li}_{3} N$ ) according to the reaction : $6 Li (s) + N_{2} (g) ⟶ 2 {Li}_{3} N (s)$
If 8.4 g of Li is taken initially with excess of $N_{2}$ (g) hen (Atomic weight of Li = 7 ) :

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Volume of nitrogen gas consumed at STP is 4.54 L

Total mass of the product obtained is 14 g

Total number of atoms obtained in the product is 1.6 N_A.

Lithium and nitrogen combine in ratio 3 : 2 by mass.

answer is A, B, C, D.

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Detailed Solution

$\underset{\frac{8.4}{7}}{6 Li} + N_{2} \to 2 {Li}_{3} N$

$1.2$ mole $\frac{1.2}{6}$ mole $\frac{1.2}{6} \times 2 \times 35 = 14 g$

$= 4.54 L at STP$

No of atoms from ${Li}_{3} N = \frac{1.2}{6} \times 2 \times 4 \times N_{A}$

$m_{L} : m_{N} :: 6 \times 7 : 14 \times 2$

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