1. limiting reactant: excess reactant:
How many complete cars can be made from these available parts?
1 Complete Race Car
only 2 Engines
available:
In chemistry, the amount of product that can be made is limited by the amount of one reactant.
Image from POGIL (Flinn HSPI) Limiting Reactants
limiting reactant:
(completely consumed)
determines the amount of product formed
excess reactant:
(left over)

2. ? N2(g) + 3 H2(g) 2 NH3(g) excess limiting “left over” “runs out”
0 H2 molecules
“left over”
“runs out”
3 mol N2 is enough to make 6 mol NH3 , but…
H2 runs out after making only 2 mol NH3 .

3. Mg + 2 HCl → MgCl2 + H2 Determining Limiting Reactant
(with Math not Pics)
Determining Limiting Reactant
convert reactant A to reactant B to compare
If available < needed (limiting)
If available > needed (excess)
Which is limiting?
Mg HCl → MgCl2 + H2
48.6 g Mg g HCl
1 mol Mg
24.31 g Mg
2 mol HCl
1 mol Mg
36.46 g HCl
1 mol HCl
48.6 g Mg x
146 g HCl x x =
(110 g HCl) available < needed (146 g HCl)
limiting
HCl is limiting
1 mol HCl
36.46 g HCl
1 mol Mg
2 mol HCl
24.31 g Mg
1 mol Mg
110. g HCl x x x =
36.7 g Mg
consumed 3

4. Mg + 2 HCl → MgCl2 + H2 Determining Limiting Reactant:
(with Math not Pics)
Determining Limiting Reactant:
convert reactant A to reactant B to compare
If available < needed (limiting)
If available > needed (excess)
Which is limiting?
Mg HCl → MgCl2 + H2
48.6 g Mg g HCl
initial
–36.7 g –110 g consumed (reacted)
11.9 g Mg g HCl left over
excess
limiting
limiting
1 mol HCl
36.46 g HCl
1 mol Mg
2 mol HCl
24.31 g Mg
1 mol Mg
110 g HCl x x x =
36.7 g Mg
consumed 4

5. Mg + 2 HCl → MgCl2 + H2 2nd Way (kind of) 48.6 g Mg 110. g HCl
1 mol MgCl2
2 mol HCl
143 g
MgCl2
produced
1 mol HCl
36.46 g HCl
95.21 g MgCl2
1 mol MgCl2
110. g HCl x x x =
1 mol MgCl2
1 mol Mg
190. g MgCl2
produced
48.6 g Mg x
1 mol Mg
24.31 g Mg
95.21 g MgCl2 1 mol MgCl2 x x =
HCl is limiting 5

6. C2H4 + 3 O2 → 2 CO2 + 2 H2O Determining Product with LR:
How many grams CO2 can be formed from
2.70 mol C2H4 and 9.30 mol O2?
C2H O2 → 2 CO H2O
3 mol O2
1 mol C2H4
2.70 mol C2H4 x =
8.10 mol O2
(9.30 mol O2) available > needed (8.10 mol O2)
O2 is excess
C2H4 is limiting
limiting
2 mol CO2
1 mol C2H4
44.01 g CO2
1 mol C2H4
238 g CO2
produced
2.70 mol C2H4 x x = 6

7. Quick Quiz!
1. How many grams of H2O can be formed from 64 g O2 and 3 mol H2?
2 H2 + O2  2 H2O
90.1 g H2O
72.1 g H2O
54.1 g H2O
36.0 g H2O
2 mol H2O
2 mol H2
18.02 g H2O
1 mol H2O
3 mol H2 x x =
1 mol O2
32 g O2
2 mol H2
1 mol O2
64 g O2 x x =
4 mol H2
(3 mol H2) available < needed (4 mol H2)
H2 is limiting