Question

question 19
1 pts
consider the following reaction: 6 li(s) + n₂(g) → 2 li₃n(s). when this reaction was run with 12.6 g of li heated with 34.4 g of n₂, 5.29 g of li₃n was formed. what is the percent yield for this reaction? report your answer to 3 significant figures.

Explanation:

Step1: Calculate moles of Li

Molar mass of Li is \(6.941\ g/mol\). Moles of Li \(n_{Li}=\frac{12.6\ g}{6.941\ g/mol}\approx1.815\ mol\)

Step2: Calculate moles of \(N_2\)

Molar mass of \(N_2\) is \(28.02\ g/mol\). Moles of \(N_2\) \(n_{N_2}=\frac{34.4\ g}{28.02\ g/mol}\approx1.228\ mol\)

Step3: Determine limiting reactant

From the balanced equation \(6Li(s)+N_2(g)\to2Li_3N(s)\), the mole ratio of \(Li\) to \(N_2\) is \(6:1\). For \(n_{N_2} = 1.228\ mol\), moles of \(Li\) required \(=6\times1.228 = 7.368\ mol\). Since we have \(1.815\ mol\) of \(Li\) (less than required), \(Li\) is the limiting reactant.

Step4: Calculate theoretical yield of \(Li_3N\)

From the balanced equation, mole ratio of \(Li\) to \(Li_3N\) is \(6:2 = 3:1\). Moles of \(Li_3N\) formed (theoretical) \(n_{Li_3N}^{theo}=\frac{1.815\ mol}{3}=0.605\ mol\)
Molar mass of \(Li_3N\) is \(34.83\ g/mol\). Theoretical yield \(m_{Li_3N}^{theo}=0.605\ mol\times34.83\ g/mol\approx21.17\ g\)

Step5: Calculate percent yield

Percent yield \(=\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{5.29\ g}{21.17\ g}\times100\%\approx25.0\%\)

Answer:

\(25.0\%\)