The correct answer is c. 0.378 g of Ca3N2
The balanced chemical equation for the reaction is:
3Ca + N2 → Ca3N2
We can see that for every 3 moles of calcium, 1 mole of nitrogen is used. Using the molar mass of calcium (40.08 g/mol) and nitrogen (28.02 g/mol) we can calculate how many moles of each reactant we have.
4.00 g of nitrogen is equal to 4.00 g / 28.02 g/mol = 0.1428 mol of nitrogen
We know that we have an excess of calcium, so we only need to consider the amount of nitrogen present.
We can use the stoichiometry of the reaction to calculate how many grams of Ca3N2 are produced.
1 mole of nitrogen reacts with 3 moles of calcium to produce 1 mole of Ca3N2
0.1428 mol N2 x (1 mole Ca3N2 / 3 mole N2) = 0.04760 mole of Ca3N2
We can convert this number of moles to grams using the molar mass of Ca3N2 (128.09 g/mol)
0.04760 mole of Ca3N2 x 128.09 g/mol = 0.378 g of Ca3N2