To determine how many grams of [tex]\( N_2 \)[/tex] must be consumed to produce 265 grams of [tex]\( CaCN_2 \)[/tex], follow these steps:
1. Calculate the molar mass of [tex]\( CaCN_2 \)[/tex]:
Given that the molar mass of [tex]\( CaCN_2 \)[/tex] is 80.11 g/mol.
2. Determine the number of moles of [tex]\( CaCN_2 \)[/tex] produced:
[tex]\[
\text{Moles of } CaCN_2 = \frac{\text{Mass of } CaCN_2}{\text{Molar mass of } CaCN_2} = \frac{265 \text{ g}}{80.11 \text{ g/mol}} \approx 3.308 \text{ mol}
\][/tex]
3. Relate moles of [tex]\( CaCN_2 \)[/tex] to moles of [tex]\( N_2 \)[/tex]:
According to the balanced chemical reaction, 1 mole of [tex]\( N_2 \)[/tex] produces 1 mole of [tex]\( CaCN_2 \)[/tex]. Therefore, the moles of [tex]\( N_2 \)[/tex] needed are the same as the moles of [tex]\( CaCN_2 \)[/tex] produced:
[tex]\[
\text{Moles of } N_2 \text{ needed} = 3.308 \text{ mol}
\][/tex]
4. Calculate the mass of [tex]\( N_2 \)[/tex] needed:
Given that the molar mass of [tex]\( N_2 \)[/tex] is 28.02 g/mol,
[tex]\[
\text{Mass of } N_2 = \text{Moles of } N_2 \times \text{Molar mass of } N_2 = 3.308 \text{ mol} \times 28.02 \text{ g/mol} \approx 92.689 \text{ g}
\][/tex]
Therefore, the reaction requires 92.689 grams of [tex]\( N_2 \)[/tex], rounded to three significant figures.
