In a combination reaction, 2.84 g magnesium is heated with 3.37 g nitrogen.?

Question

A) Which reactant is limiting?
B) How many grams of product are formed?

Answer 1

Step 1: Write out a equation for your reaction:
Initial: Mg + N2 = MgN
Mg's oxidation state is 2, N is +/-3. The most likely combination is therefore Mg3N2 (which is magnesium nitride)
Final: 3Mg + N2 = Mg3N2

Thus, you need 3 moles of Mg per 2 moles of Nitrogen.
N has a molar mass of 14g
Mg has a molar mass of 24.3050g
2.84g Mg / 24.3050 g/mol = 0.1168 moles Mg
3.37g N / 14 g/mol = 0.2407 moles N.

a.) Since you need 3 moles Mg per 2 moles N, Mg is the limiting reactant. (solution!)

3 moles of Mg create 1 mole Mg3N2.
If 0.1168 moles of Mg are reacting, it will create 0.1168 / 3 moles or 0.038933... moles of Mg3N2.
Mg3N2 has a molar mass of 14 * 2 + 24.3050 * 3 = 100.915 g/mol.
0.038933 mol * 100.915 g/mol = 3.93 g Mg3N2 created. (solution for b).