Question

what is the theoretical yield, in grams, of li₃n when the reaction begins with 4.32 moles of li and an excess of nitrogen?
6 li(s) + n₂(g) → 2 li₃n(s)
o 50.2 g li₃n
o 453 g li₃n
o 151 g li₃n
o 301 g li₃n

Explanation:

Step1: Determine mole ratio

From the balanced equation \(6Li(s)+N_2(g)\to2Li_3N(s)\), the mole ratio of \(Li\) to \(Li_3N\) is \(6:2 = 3:1\).

Step2: Calculate moles of \(Li_3N\)

Given \(n(Li)=4.32\) moles. Using the mole ratio, \(n(Li_3N)=\frac{4.32}{3}=1.44\) moles.

Step3: Calculate molar mass of \(Li_3N\)

Molar mass of \(Li_3N\): \(M = 3\times6.94 + 14.01=34.83\space g/mol\).

Step4: Calculate mass of \(Li_3N\)

Using \(m = n\times M\), \(m(Li_3N)=1.44\times34.83 = 50.1552\approx50.2\space g\).

Answer:

\(50.2\space g\space Li_3N\) (the option \(50.2\space g\space Li_3N\) is correct)