Question

question 16
when heated, lithium reacts with nitrogen to form lithium nitride according to the following reaction: 6 li(s) + n2(g) → 2 li3n(s). what is the theoretical amount of lithium nitride that can be made, in g, if one begins with 12.6 g of li and 34.4 g of n2? report your answer to 3 significant figures.

Explanation:

Step1: Calculate moles of Li and N₂

• Molar mass of Li: \(M_{Li}=6.94\ g/mol\)
• Moles of Li: \(n_{Li}=\frac{m_{Li}}{M_{Li}}=\frac{12.6\ g}{6.94\ g/mol}\approx1.82\ mol\)
• Molar mass of N₂: \(M_{N_2}=28.02\ g/mol\)
• Moles of N₂: \(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{34.4\ g}{28.02\ g/mol}\approx1.23\ mol\)

Step2: Determine limiting reactant

From the balanced equation \(6Li(s)+N_2(g)\to2Li_3N(s)\), the mole ratio of Li to N₂ is \(6:1\).

• For \(n_{Li} = 1.82\ mol\), moles of N₂ required: \(n_{N_2,\text{required}}=\frac{1.82\ mol}{6}\approx0.303\ mol\)
• Since \(n_{N_2,\text{available}} = 1.23\ mol>0.303\ mol\), Li is the limiting reactant.

Step3: Calculate moles of Li₃N

Using the mole ratio of Li to Li₃N (\(6:2 = 3:1\))

• Moles of Li₃N: \(n_{Li_3N}=\frac{n_{Li}}{3}=\frac{1.82\ mol}{3}\approx0.607\ mol\)

Step4: Calculate mass of Li₃N

Molar mass of Li₃N: \(M_{Li_3N}=(3\times6.94 + 14.01)\ g/mol=(20.82+14.01)\ g/mol = 34.83\ g/mol\)

• Mass of Li₃N: \(m_{Li_3N}=n_{Li_3N}\times M_{Li_3N}=0.607\ mol\times34.83\ g/mol\approx21.2\ g\)

Answer:

\(21.2\ g\)