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