QuestionSeptember 20, 2025

Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li(s)+N_(2)(g)arrow 2Li_(3)N(s) How many moles of lithium are needed to produce 21.0 mol of Li_(3)N when the reaction is carried out in the presence of excess nitrogen? 42.0 mol 63.0 mol 7.00 mol 14.0 mol 21.0 mol

Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li(s)+N_(2)(g)arrow 2Li_(3)N(s) How many moles of lithium are needed to produce 21.0 mol of Li_(3)N when the reaction is carried out in the presence of excess nitrogen? 42.0 mol 63.0 mol 7.00 mol 14.0 mol 21.0 mol
Lithium and nitrogen react in a combination reaction to produce lithium nitride:
6Li(s)+N_(2)(g)arrow 2Li_(3)N(s)
How many moles of lithium are needed to produce 21.0 mol of Li_(3)N when the reaction is
carried out in the presence of excess nitrogen?
42.0 mol
63.0 mol
7.00 mol
14.0 mol
21.0 mol

Solution
4.4(227 votes)

Answer

63.0 mol Explanation 1. Write the mole ratio from the balanced equation From 6Li + N_2 \rightarrow 2Li_3N, the mole ratio is 6:2 or 3:1 for Li : Li_3N. 2. Set up proportion to solve for moles of Li 3 mol Li produces 1 mol Li_3N. For 21.0 mol Li_3N: x = 3 \times 21.0 = 63.0 mol Li.

Explanation

1. Write the mole ratio from the balanced equation<br /> From $6Li + N_2 \rightarrow 2Li_3N$, the mole ratio is $6:2$ or $3:1$ for $Li : Li_3N$.<br />2. Set up proportion to solve for moles of Li<br /> $3$ mol $Li$ produces $1$ mol $Li_3N$. For $21.0$ mol $Li_3N$: $x = 3 \times 21.0 = 63.0$ mol $Li$.
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