Question
When 100.g Mg3N2 reacts with 75.0 g H2O, 15.0 g of NH3 is formed. What is the % yield?
Answers
R_scott
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)
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)
DrBob222
% yield ... 100*[15.0 g / (1.39 mol * 17.0 g/mol)]