To determine the percent yield of calcium nitride ([tex]\( Ca_3N_2 \)[/tex]) from the reaction of calcium ([tex]\( Ca \)[/tex]) with nitrogen gas ([tex]\( N_2 \)[/tex]), we need to follow these steps:
1. Calculate the moles of calcium (Ca) used:
[tex]\[
\text{Molar mass of Ca} = 40.08 \, \text{g/mol}
\][/tex]
[tex]\[
\text{Mass of Ca} = 58.6 \, \text{g}
\][/tex]
To find the moles of [tex]\( Ca \)[/tex], we use the formula:
[tex]\[
\text{Moles of Ca} = \frac{\text{Mass of Ca}}{\text{Molar mass of Ca}} = \frac{58.6 \, \text{g}}{40.08 \, \text{g/mol}} = 1.4620758483033933 \, \text{moles}
\][/tex]
2. Determine the theoretical moles of [tex]\( Ca_3N_2 \)[/tex] produced:
According to the balanced chemical equation:
[tex]\[
3 \, Ca (s) + N_2 (g) \rightarrow Ca_3N_2 (s)
\][/tex]
From stoichiometry, 3 moles of [tex]\( Ca \)[/tex] produce 1 mole of [tex]\( Ca_3N_2 \)[/tex]. Thus:
[tex]\[
\text{Moles of } Ca_3N_2 = \frac{\text{Moles of } Ca}{3} = \frac{1.4620758483033933}{3} = 0.4873586161011311 \, \text{moles}
\][/tex]
3. Calculate the theoretical mass of [tex]\( Ca_3N_2 \)[/tex] that should be produced:
[tex]\[
\text{Molar mass of } Ca_3N_2 = 3 \times 40.08 \, \text{g/mol} + 2 \times 14.01 \, \text{g/mol} = 148.256 \, \text{g/mol}
\][/tex]
[tex]\[
\text{Theoretical mass of } Ca_3N_2 = \text{Moles of } Ca_3N_2 \times \text{Molar mass of } Ca_3N_2 = 0.4873586161011311 \, \text{moles} \times 148.256 \, \text{g/mol} = 72.2557884231537 \, \text{g}
\][/tex]
4. Calculate the percent yield:
[tex]\[
\text{Actual mass of } Ca_3N_2 \text{ produced} = 32.4 \, \text{g}
\][/tex]
[tex]\[
\text{Percent yield} = \left(\frac{\text{Actual mass of } Ca_3N_2}{\text{Theoretical mass of } Ca_3N_2}\right) \times 100\% = \left(\frac{32.4 \, \text{g}}{72.2557884231537 \, \text{g}}\right) \times 100\% = 44.84069817390259\%
\][/tex]
Therefore, the percent yield of [tex]\( Ca_3N_2 \)[/tex] is [tex]\( 44.84\% \)[/tex].
