Final answer:
To produce 1 kilogram of calcium nitride without leftovers, 811.63 grams of calcium and 188.86 grams of nitrogen are needed, based on stoichiometric calculations and the compound's molar mass.
Step-by-step explanation:
The question is centered on a stoichiometry problem, specifically determining the correct amounts of calcium and nitrogen needed to produce a certain mass of calcium nitride without any waste.
First, the molar mass of calcium nitride (Ca3N2) must be calculated. The molar mass of calcium is approximately 40.08 g/mol and for nitrogen, it is approximately 14.01 g/mol. Calcium nitride contains 3 moles of calcium and 2 moles of nitrogen. Thus, the molar mass of calcium nitride is (3 × 40.08 g) + (2 × 14.01 g) = 120.24 g + 28.02 g = 148.26 g/mol.
Now, we need to find the amount of calcium and nitrogen necessary to produce 1.0 kg (1000 g) of calcium nitride without any excess reactants. This requires a stoichiometric calculation based on the mole ratio from the balanced chemical equation for the reaction forming calcium nitride: 3Ca + N2 → Ca3N2. Since the molar mass of calcium nitride is 148.26 g/mol, 1.0 kg (or 1000 g) of Ca3N2 equates to 1000 g / 148.26 g/mol ≈ 6.744 moles of Ca3N2.
For every mole of Ca3N2 produced, 3 moles of calcium and 1 mole of nitrogen gas are required. Consequently, to create 6.744 moles of Ca3N2, we need 3 × 6.744 moles of calcium and 6.744 moles of nitrogen gas. Therefore, 3 × 6.744 × 40.08 g/mol of calcium and 6.744 × 28.02 g/mol of nitrogen are needed. This results in 811.63 grams of calcium and 188.86 grams of nitrogen required to produce 1 kilogram of calcium nitride with no leftover reactants.