When 100.g Mg3N2 reacts with 75.0 g H2O, 15.0 g of NH3 is formed. What is the % yield?

Answer 1

Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g)

water is the limiting reactant ... 75.0 g / 18.0 g/mole = 4.17 moles

expected NH3 ... 4.17 / 3 = 1.39 moles
... molar mass NH3 ... 17.0 g

% yield ... 15.0 g / (1.39 mol * 17.0 g/mol)

Answer 2

% yield ... 100*[15.0 g / (1.39 mol * 17.0 g/mol)]