Other way to identify the Limiting Reactant: Calculate the Experimental mole ratio and the theoretical mole ratio, and then, compare both.

Experimental data: N2= 3.2 mol & O2= 4.8 mol Determine the limiting reactant if you are combining 3.2 mol of N2 with 4.8 mol of O2. 2 N2(g) + 5 O2(g) 2 N2O5(g) Experimental data: N2= 3.2 mol & O2= 4.8 mol Mole ratios: Stoichiometric = 2/5 = 0.4 (N2 /O2) Experimental = 3.2/4.8 = 0.67 Experimental > Stoichiometric Limiting reactant Excess

Mole ratios: Experimental > Stoichiometric Numerator in excess Denominator limiting Numerator limiting Denominator in excess Experimental < Stoichiometric No excess & no limiting substances Experimental = Stoichiometric

Exercise: Determine the mass of hydrogen formed when 100 g of magnesium combined with 100 g of hydrochloric acid. Mg(s) +2 HCl(aq) → MgCl2(aq) + H2(g)

Exercise: Determine the mass of hydrogen formed when 100 g of magnesium combined with 100 g of hydrochloric acid 100 g 100 g ? g Mg(s) +2 HCl(aq) → MgCl2(aq) + H2(g)

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) Mg = 100 g ÷ 24 g/mol = 4.17 mol HCl = 100 g ÷ 36.5 g/mol = 2.74 mol Stoichiometric = 1/2 = 0.5 Mole ratios: (Mg / HCl) Experimental = 4.17/2.74 =1.52 Experimental > Stoichiometric Limiting reactant Excess

? g Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) m(Mg) = 100 g m(HCl) = 100 g (Limiting Reactant) = 2.74 mol = = 1.37 mol H2 X = m (H2) = 1.37 mol H2 x 2.02 g/mol = 2.77 g H2 How much of the excess reactant left over?

m(Mg) = 100 g = 4.17 mol (Excess reactant) How much of the excess reactant left over? m(Mg) = 100 g = 4.17 mol (Excess reactant) m(HCl) = 100 g (Limiting Reactant) = 2.74 mol Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) X mol 2.74 mol 1 mol 2 mol X = = X = 1.37 mol Mg (reacted) 4.17 mol – 1.37 mol = 2.8 mol Excess of Mg(s) m(Mg) = 2.8 mol x 24 g/mol = 67.2 g Mg