Answer:
About 10.2 g Li₃N.
Explanation:
We are given the reaction:
[tex]\displaystyle \text{6 Li(s) + N$_2$(g) }\longrightarrow \text{2 Li$_3$N(s)}[/tex]
And we want to determine the amount of Li₃N that can be formed from 6.07 g of Li and an excess of nitrogen.
To convert from g Li to g Li₃N, we can: (1) convert from g Li to mol Li, (2) mol Li to mol Li₃N, and (3) mol Li₃N to g Li₃N.
Molecular weight of Li₃N:
[tex]\displaystyle \begin{aligned}\text{MW}_\text{Li$_3$N} & = (3 (6.94) + 14.01) \text{ g/mol} \\ \\ & =34.83\text{ g/mol} \end{aligned}[/tex]
This yields three ratios:
[tex]\displaystyle \frac{1 \text{ mol Li}}{6.94 \text{ g Li}}, \frac{2 \text{ mol Li$_3$N}}{6 \text{ mol Li}}, \text{ and } \frac{34.83 \text{ g Li$_3$N}}{1 \text{ mol Li$_3$N}}[/tex]
From the initial value, multiply:
[tex]\displaystyle 6.07 \text{ g Li} \cdot \frac{ 1 \text{ mol Li}}{6.94 \text{ g Li}} \cdot \frac{2 \text{ mol Li$_3$N}}{6 \text{ mol Li}} \cdot \frac{34.83 \text{ g Li$_3$N}}{1 \text{ mol Li$_3$N}} = 10.2\text{ g Li$_3$N}[/tex]
In conclusion, 10.2 g of Li₃N is formed from 6.07 g of Li and an excess of nitrogen.
