Question

$Li$ metal is one of the few substances that reacts directly with molecular nitrogen. The balanced equation for reaction is: $6Li\left(s\right)+N_2\left(g\right)\to 2L{i}_{3}N\left(s\right)$
How many grams of the product, lithium nitride, can be prepared from $3.5g$ of iithium metal and $8.4g$ of molecular nitrogen?

• A. $21.00g$ of $L{i}_{3}N$
• B. $2.91g$ of $L{i}_{3}N$
• C. $5.83g$ of $L{i}_{3}N$
• D. $10.50g$ of $L{i}_{3}N$

Answer 1

Answer: (C)

$5.83g$ of $L{i}_{3}N$

$6Li+N_2⟶2L{i}_{3}N$

Moles:

$n_{Li}=\frac{3.5}{7}=0.5$ $moles$

$n_{N_2}=\frac{8.4}{28}=0.3$ $moles$

$1$ $mole$ $N_2$ requires $6$ $mole$ $Li$

$\therefore 0.3$ $mole$ $N_2$ will need $6\times 0.3=1.8$ $mole$ $Li$

But we have only $0.5$ $mole$ $Li$, hence it will get over first.

$Li$ is the limiting reagent.

$6$ $mole$ of $Li$ produces $2$ $mole$ $L{i}_{3}N$

$\therefore 0.5$ $mole$ of $Li$ produces $x$ $mole$ $L{i}_{3}N$

$x=\frac{0.5\times 2}{6}=\frac{1}{6}$ $mole$

Mass of $L{i}_{3}N=$ Moles $\times$ Molar mass of $L{i}_{3}N=\frac{1}{6}\times 35=5.83$ $grams$