12.3 Limiting Reactants


Limiting reactant: runs out first; determines the amount of product made
Excess reactant: left-over reactants
How to determine the limiting reactant: the reactant that gives less product when calculated
How to determine the remaining excess reactant: use the amount of product made to determine the amount of
reactant used, then subtract the amount used from the amount at the start of the reaction
EX: problem 12-5 p. 367
The reaction between solid white phosphorus and oxygen produces solid tetraphosphorus decoxide (P4O10).
This compound is often called diphosphorus pentoxide because its empirical formula is P2O5.
a) Determine the mass of P4O10 formed if 25.0 g of phosphorus (P4) and 50.0 g of oxygen are combined.
b) How much of the excess reactant remains after the reaction occurs?
Step 1: write & balance the chemical equation
P4 + 5O2 → P4O10
Step 2: Calculate the molar masses of both knowns and the unknown
P: 4  30.974  123.896 g/mol P4
O: 2 15.999  31.998 g/mol O2
P: 4  30.974  123.896
O: 10 15.999  159.99
sum = 283.89 g/mol P4O10
Step 3: Write the mole ratios you will need (one for each reactant)
1 mol P4 O10
1 mol P4
1 mol P4 O10
5 mol O 2
Step 4: do mass-to-mass calculations for both reactants
25.0 g P4 
1 mol P4 O10 283.89 g P4O10
1 mol P4


 57.3 g P4O10
123.896 g P4
1 mol P4
1 mol P4O10
50.0 g O 2 
1 mol P4 O10 283.89 g P4O10
1 mol O2


 444 g P4O10
31.998 g O 2 1 mol O 2
1 mol P4O10
a) The limiting reactant is P4. The actual amount of P4O10 produced is 57.3 g.
Step 5: Use the calculated mass of product as your known and calculate the mass of the excess reactant used
57.3 g P4 O10 
1 mol P4 O10
5 mol O 2 31.998 g O 2


 32.3 g O 2
283.89 g P4O10 1 mol P4O10
1 mol O 2
Step 6: Subtract the calculated mass of excess reactant from the given mass of it
b) 50.0 – 32.3 = 17.7 g O2 leftover after the reaction
EX: p. 368 #21a-c
Knowns: 88.0 g CO2 & 64.0 g H2O
Unknown: grams C6H12O6
Step 1: 6CO2 + 6H2O → C6H12O6 + 6O2
Step 2:
Step 3:
unknown: H 2 O
unknown: CO2
H 2  1.008 = 2.016
O 1  15.999 = 15.999
C 1  12.011 = 12.011
O 2  15.999 = 31.998
sum = 18.015 g/mol H 2 O
sum = 44.009 g/mol CO2
1 mol C6 H12O6
6 mol CO 2
Step 4: 88.0 g CO 2 
64.0 g H 2O 
known: C6 H12O6
C 6  12.011 = 72.066
H 12  1.008 = 12.096
O 6  15.999 = 95.994
sum = 180.156 g/mol C6 H12 O6
1 mol C6 H12O6
6 mol H 2O
1 mol C6 H12O6 180.156 g C6 H12O 6
1 mol CO 2


 60.0 g C6 H12O 6
44.009 g CO 2
6 mol CO 2
1 mol C6 H12O 6
1 mol H 2 O 1 mol C6 H12 O6 180.156 g C6 H12O6


 107 g C6 H12O 6
18.015 g H 2O
6 mol H 2O
1 mol C6 H12O 6
a) The limiting reactant is CO2.
Step 5: 60.0 g C6 H12O 6 
1 mol C6 H12O6
6 mol H 2O
18.015 g H 2O


 36.0 g H 2O
180.156 g C6 H12O 6 1 mol C 6 H12O 6
1 mol H 2O
64.0 – 36.0 = 28.0
b) The excess reactant is H2O and there will be 28.0 g in excess.
c) The mass of glucose produced is 60.0 g.