To determine how many grams of [tex]\( N_2 \)[/tex] must be consumed to produce 265 grams of [tex]\( CaCN_2 \)[/tex], we will follow these steps:
1. Determine the molar mass of [tex]\( CaCN_2 \)[/tex] and [tex]\( N_2 \)[/tex]:
- The molar mass of [tex]\( CaCN_2 \)[/tex] is given as 80.1 g/mol.
- The molar mass of [tex]\( N_2 \)[/tex] is given as 28.0 g/mol.
2. Calculate the moles of [tex]\( CaCN_2 \)[/tex] produced:
- Mass of [tex]\( CaCN_2 \)[/tex] produced = 265 grams.
- To find the moles of [tex]\( CaCN_2 \)[/tex], use the formula:
[tex]\[
\text{Moles of } CaCN_2 = \frac{\text{Mass of } CaCN_2}{\text{Molar mass of } CaCN_2}
\][/tex]
[tex]\[
\text{Moles of } CaCN_2 = \frac{265 \, \text{g}}{80.1 \, \text{g/mol}} \approx 3.308
\][/tex]
3. Relate the moles of [tex]\( CaCN_2 \)[/tex] to the moles of [tex]\( N_2 \)[/tex]:
- From the balanced equation, 1 mole of [tex]\( N_2 \)[/tex] produces 1 mole of [tex]\( CaCN_2 \)[/tex].
- Therefore, the moles of [tex]\( N_2 \)[/tex] required are equal to the moles of [tex]\( CaCN_2 \)[/tex] produced:
[tex]\[
\text{Moles of } N_2 = \text{Moles of } CaCN_2 \approx 3.308
\][/tex]
4. Calculate the mass of [tex]\( N_2 \)[/tex] required:
- To find the mass of [tex]\( N_2 \)[/tex], use the formula:
[tex]\[
\text{Mass of } N_2 = \text{Moles of } N_2 \times \text{Molar mass of } N_2
\][/tex]
[tex]\[
\text{Mass of } N_2 = 3.308 \, \text{moles} \times 28.0 \, \text{g/mol} \approx 92.634 \, \text{g}
\][/tex]
So, the reaction requires approximately 92.634 grams of [tex]\( N_2 \)[/tex]. Rounding to three significant figures, the answer is:
The reaction requires [tex]\( 92.6 \)[/tex] grams of [tex]\( N_2 \)[/tex].
