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

2. 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) O2(g) 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

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

4. 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)

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

6. 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

7. ? 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?

8. 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