1. _____ Cu(s) + _____ AgNO3(aq) → _____ Cu(NO3)2(aq) + _____ Ag(s)
If 2.5 moles of copper and 5.5 moles of silver nitrate are available to react, what is the limiting
reactant?
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
2 mol Ag
2.5 mol Cu × 1 mol Cu = 5 mol Ag possible from 2.5 mol Cu
2 mol Ag
5.5 mol AgNO3 × 2 mol AgNO3 = 5.5 mol Ag possible from 5.5 mol AgNO3 so Cu is the LR
One can compare the other reactant, with the same result. This is not necessary, but is illustrated below:
2.5 mol Cu ×
1 mol Cu(NO3)3
= 2.5 mol Cu(NO3)3 possible from 2.5 mol Cu
1 mol Cu
5.5 mol AgNO3 ×
1 mol Cu(NO3)3
2 mol AgNO3 = 2.75 mol Cu(NO3)3 possible from 5.5 mol AgNO3 so Cu is
LR when looking at the other product. This is always true.
2. CaO(s) + H2O(l) → Ca(OH)2(aq)
How many grams of calcium hydroxide will be formed in this reaction when 4.44 g of calcium oxide
and 7.77 g of water are available to react? Also identify the limiting and excess reactants.
4.44 g CaO ×
1 mol CaO
1 mol Ca(OH)2 74.09 g Ca(OH)2
×
×
= 5.87 g Ca(OH)2 possible from CaO
1 mol CaO
1 mol Ca(OH)2
56.08 g CaO
7.77 g H2O ×
1 mol H2O
1 mol Ca(OH)2 74.09 g Ca(OH)2
×
×
= 31.95 g Ca(OH)2 possible from H2O
1 mol H2O
1 mol Ca(OH)2
18.02 g H2O
So 5.87 g 5.87 g Ca(OH)2 is the maximum, theoretical yield, from CaO as the Limiting Reactant, and H2O is in
excess.
3. Magnesium undergoes a single displacement reaction with hydrochloric acid (aq) to produce
hydrogen gas and magnesium chloride (aq).
Balanced Equation: Mg(s) + 2HCl(aq) → MgCl2(g) + H2(g) (H is diatomic, Horses)
How many grams of hydrogen gas will be produced from the reaction of 3.00 g of magnesium with
4.00 g of hydrochloric acid?
7.77 g HCl ×
3.00 g Mg ×
1 mol HCl
1 mol H2 2.02 g H2
×
×
= 0.249 g H2 possible from HCl
36.45 g HCl 2 mol HCl 1 mol H2
1 mol Mg
1 mol H2 2.02 g H2
×
×
= 0.111 g H2 possible from Mg
24.31 g Mg 1 mol Mg 1 mol H2
So Mg is the limiting reactant (LR) with 0.111 g of H2 forming (theoretical yield) and HCl is in excess.
Identify the limiting and excess reactants.
4. Sulfur reacts with oxygen gas to produce sulfur trioxide gas.
Balanced Equation:
2S(s) + 3O2(g) → 2SO3(g)
If 6.3 g of sulfur reacts with 10.0 g of oxygen gas, what is the limiting reactant?
How many grams of sulfur trioxide will be produced?
(I added an extra digit to the 6.3 g of sulfur to even up the sig figs)
6.30 g S ×
1 mol S
2 mol SO3 80.07 g SO3
×
×
= 15.7 g SO3 possible from S
2 mol S
1 mol SO3
32.07 g S
10.0 g O2 ×
1 mol O2
2 mol SO3 80.07 g SO3
×
×
= 16.7 g SO3 possible from O2
1 mol SO3
32.00 g O2 3 mol O2
So, the sulfur is the LR and O2 is in excess, with 15.7 g of SO3 as the theoretical yield.