Explanation:
\(6Li(s)+N_2(g)\rightarrow 2Li_3N(s)\)
1. Mass of \(N_2\) needed to react with 0.536 moles of Li.
According to reaction, 6 moles of Li reacts with 1 mol of \(N_2\).
Then 0.536 moles of Li will react with:
\(\frac{1}{6}\times 0.536\) moles of \(N_2\) that is 0.0893 moles.
Mass of \(N_2[tex] gas needed:
[tex]=28 g/mol\times 0.0893 mol=2.5004 g\)
2.The number of moles of Li required to make 46.4 g of \(Li_3N\)
Moles of \(Li_3N=\frac{46.4 g}{35 g/mol}=1.3257 mol\)
According to reaction the 2 moles of \(Li_3N\) are produced from 6 moles of Li.
Then 1.3257 moles of \(Li_3N\) will produced from:
\(\frac{6}{2}\times 1.3257=3.9771 moles\)
3.9771 moles of lithium will needed.
3. The mass in grams of \(Li_3N\) produced from 3.65 g Li.
Moles of Li \(=\frac{3.65 g}{7 g/mol}=0.5214 moles\)
According to reaction, 6 moles of Li gives 2 moles of \(Li_3N\)
Then 0.5214 moles of Li will give \(\frac{2}{6}\times 0.5214\) that is 0.1738 moles of \(Li_3N\).
Mass of\(Li_3N=0.1738 mole\times 35 g/mol=6.083 g\)
6.083 grams of \(Li_3N\) will be produced.
4. The number of moles of lithium needed to react with 7.00 grams of \(N_2\).
Moles of \(N_2=\frac{7.00 g}{28 g/mol}=0.25 moles\)
1 mol of \(N_2\) reacts with 6 mol of Li
Then, 0.25 moles of \(N_2\) will ftreact with :
\(6\times 0.25 moles=1.5 moles\) of lithium
1.5 moles of Li will be needed.
