Lithium and nitrogen react in a combination reaction to produce lithium nitride:
6Li(s)+ $\mathrm { N } _ { 2 }$ (g)? $2 \mathrm { Li } _ { 3 }$ N(s)
How many moles of lithium are needed to produce 0.40 mol of $\mathrm { Li } _ { 3 }$ N when the reaction is carried out in the presence of excess nitrogen?

Question

Lithium and nitrogen react in a combination reaction to produce lithium nitride: 
6Li(s)+ $\mathrm { N } _ { 2 }$ (g)? $2 \mathrm { Li } _ { 3 }$ N(s)
How many moles of lithium are needed to produce 0.40 mol of $\mathrm { Li } _ { 3 }$ N when the reaction is carried out in the presence of excess nitrogen?

Quizplus · Accepted Answer

From the balanced chemical equation, we see that 6 moles of lithium react to produce 2 moles of lithium nitride:
6Li(s) + N2(g) → 2Li3N(s)
Thus, the ratio of lithium to lithium nitride is 6:2 or 3:1. Therefore, to produce 0.40 moles of lithium nitride, we need 0.40/3 = 0.1333... moles of lithium. The answer closest to this value is B) 1.2, which represents 1.2/6 or 0.2 moles of lithium.

Lithium and Nitrogen React in a Combination Reaction to Produce N2\mathrm { N } _ { 2 }N2​

Lithium and Nitrogen React in a Combination Reaction to Produce $\mathrm { N } _ { 2 }$