Question

when heated, lithium reacts with nitrogen to form lithium nitride:
6li(s) + n₂(g) → 2li₃n(s)
part: 0 / 2
part 1 of 2
what is the theoretical yield of li₃n in grams when 12.5 g of li is heated with 34.1 g of n₂? round your answer to 3 significant digits.
g

Explanation:

Step1: Calculate moles of Li and \(N_2\)

• Molar mass of \(Li\): \(M_{Li}=6.94\ g/mol\). Moles of \(Li\), \(n_{Li}=\frac{m_{Li}}{M_{Li}}=\frac{12.5\ g}{6.94\ g/mol}\approx1.80\ mol\)
• Molar mass of \(N_2\): \(M_{N_2} = 28.02\ g/mol\). Moles of \(N_2\), \(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{34.1\ g}{28.02\ g/mol}\approx1.22\ mol\)

Step2: Determine the limiting reactant

From the balanced equation \(6Li(s)+N_2(g)\to2Li_3N(s)\), the mole ratio of \(Li\) to \(N_2\) is \(6:1\).
If \(n_{N_2} = 1.22\ mol\), the moles of \(Li\) required for complete reaction with \(N_2\) is \(n_{Li,\ required}=6\times n_{N_2}=6\times1.22 = 7.32\ mol\). But we have \(n_{Li}=1.80\ mol\).
If \(n_{Li}=1.80\ mol\), the moles of \(N_2\) required for complete reaction with \(Li\) is \(n_{N_2,\ required}=\frac{n_{Li}}{6}=\frac{1.80\ mol}{6}=0.3\ mol\). Since \(n_{N_2}=1.22\ mol>0.3\ mol\), \(Li\) is the limiting reactant.

Step3: Calculate moles of \(Li_3N\)

From the balanced equation, mole ratio of \(Li\) to \(Li_3N\) is \(6:2 = 3:1\). So moles of \(Li_3N\), \(n_{Li_3N}=\frac{n_{Li}}{3}=\frac{1.80\ mol}{3}=0.6\ mol\)

Step4: Calculate mass of \(Li_3N\)

Molar mass of \(Li_3N\): \(M_{Li_3N}=(3\times6.94 + 14.01)\ g/mol=(20.82+14.01)\ g/mol = 34.83\ g/mol\)
Mass of \(Li_3N\), \(m_{Li_3N}=n_{Li_3N}\times M_{Li_3N}=0.6\ mol\times34.83\ g/mol = 20.9\ g\)

Answer:

\(20.9\ g\)