When 58.6 g of calcium reacts with nitrogen gas, 32.4 g of calcium nitride is produced.

Part A:
What is the percent yield of calcium nitride for this reaction?

[tex]\[ 3 \text{Ca} (s) + \text{N}_2 (g) \rightarrow \text{Ca}_3\text{N}_2 (s) \][/tex]

Express your answer with the appropriate units.

The percent yield of [tex]\(\text{Ca}_3\text{N}_2\)[/tex] = [tex]\(\boxed{\text{Value}}\)[/tex] [tex]\(\boxed{\text{Units}}\)[/tex]



Answer :

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].

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