# Limiting Reactants

```Limiting Reactants
It will be necessary to determine the limiting reactant whenever we are
given the amounts of two or more reactants in a chemical reaction. The
limiting reactant limits the amount of product that can be obtained and
is therefore totally consumed in the reaction. Any other reactant is in
excess.
Problem
I:
mole
Given:
9.25 moles of potassium chloride and 8.75 moles of chromium (III) chloride
react according to the following equation:
mole
2 KCl(aq) + 2 CrCl3(aq)
7H2O(l) + 3 Cl2(g)
Problem:
1. find the limiting reactant
2. Find the amount of chlorine formed
3. Find the amount of excess
Solution:
1. Compare moles of each reactant to moles of product (Cl2)
2. The amount of chlorine formed is 13.1 mol.
3. Amount of excess used (Look back at the original moles of reactants.)
n = 8.75 moles KCl(aq) used up
Excess = original moles KCl(aq) - moles used KCl(aq)
Excess = 9.25 moles - 8.75 moles
Excess = 0.50 moles
Problem II: Volume/concentration
moles
mass
Given:
348 mL of 1.50
Ba(OH)2(aq) is added to 285 mL of 2.30
and reacts according to the following equation:
3Ba(OH)2(aq) + 2H3P04(aq)
Ba3(P04)2(aq) + 6H2O(l)
Problem:
1. Find the limiting reactant.
2. Find the mass of barium phosphate formed.
3. Find the mass of reagent in excess.
Solution:
1. 3Ba(OH)2(aq) + 2H3P04(aq)
of H3PO4(aq)
Ba3(P04)2(aq) + 6H2O(l)
V = 348 mL = 0.348 L
V = 285 mL = 0.285 L
C = 1.50
n=CV
C = 2.30
n=CV
n = (1.50
)(0.348 L )
n = 0.522 mol
n = (2.30
)(0.285 L)
n = 0.6555 mol
n = 0.32775 mol
This is the smaller amount,
therefore, Ba(OH)2(aq) is the limiting reactant
2. The mass of Ba3(P04)2(aq) formed is 0.1 74 mol x
3. Excess used up
n = 0.116 mol
Excess = original moles - moles used up of H3P04(aq)
Excess = 0.6555 mol - 0.116 mol
Excess = 0.5395 mol
Excess = 0.5395 mol x
Method gram
I
mole
gram
Given:
4.20 grams of sulphuric acid and 7.30 grams of sodium hydroxide react
according to the equation:
H2SO4(aq) +2NaOH(aq)
Problem:
Na2 S04(aq) + 2H2O(l)
1. Find the limiting reactant.
2. Find the amount of product formed. Na2 S04(aq)
3. Find the amount of excess.
Solution: H2SO4(aq) +2NaOH(aq)
Na2 S04(aq) + 2H2O(l)
1.
a.
b. compare mole of reactants to moles of product
This is the smaller amount, therefore,
H2SO4(aq) is the limiting reactant.
2.
a. The amount of product formed is 4.28 x 10-2 moles Na2 S04(aq)
b. Amount of product in grams is 4.28 x 10-2 moles x 142.04/mol =
6.08 g
3.
a. Amount of excess used (go back to original moles and look at
reactants)
x = 8.56 x 10-2 moles NaOH(aq) used
b. Excess = original moles NaOH(aq) - molesNaOH(aq) used
Excess = 1.83 x 10-1 moles - 8.56 x 10-2 moles
Excess = 9.74 x 10-2 moles NaOH(aq)
c. Amount of excess in grams 9.74 x 10-2 moles x
NaOH(aq)
= 3.90 g
Method
II
gram
Given:
4.20 grams of sulphuric acid and 7.30 grams of sodium hydroxide react
according to the equation:
gram
H2SO4(aq) +2NaOH(aq)
Problem:
Na2 S04(aq) + 2H2O(l)
1. Find the limiting reactant.
2. Find the amount of product formed. Na2 S04(aq)
3. Find the amount of excess.
Solution:
compare each of the reactants to the product
1. This is the smaller amount, therefore,
H2SO4(aq) is the limiting reactant.
2. The amount of product formed is 6.08 g.
3.
a. Amount of excess (go back to original masses and look at the
reactants).
x = 3.43 g (NaOH used up)
b. Excess = original mass - mass used up
Excess = 7.30 g - 3.43 g
Excess = 3.90 g
Using the following equation, calculate the concentration of each after the solutions are
mixed.
a.
b.
c.
[Ca2+(aq)]
[NO3-(aq)]
[Ag+(aq)]
CaCl2(aq) + 2AgNO3(aq)
2AgCl(s) + Ca(N03)2(aq)
V = 25.0 mL
V = 20.0 mL
C = 1.78
C = 6.50
n = 0.0445 mol
n = 0.130 mol
Ca2+(aq) + 2NO3-(aq)
Ca(NO3)2(aq)
[Ag+(aq)] Since AgNO3(aq) is in excess part of the Ag will go into solution as ions, the rest
will form the solid AgCl(s). Find the amount used up and the excess.
n = 0.0890 mol AgNO3(aq) used up
Excess = 0.130 mol - 0.089 mol = 0.0410 mol
Exercises
A. CaCl2(aq) + 2AgNO3(aq)
2AgCl(s) + Ca(N03)2(aq)
Given: 4.32 moles of calcium chloride react with 8.75 moles of silver nitrate.
Calculate the amount of silver chloride produced.
B. 2C4H10(g) + 1302(g)
8CO2(g) + 10H2 O(g)
Given: 20.0 L of butane reacts with 90.0 L of oxygen gas at STP.
Find the volume of carbon dioxide gas formed.
C. 2NaI(aq) + Pb(NO3)2(aq)
PbI2(s) + 2NaNO3(aq)
sodium iodide reacts with 4.58 L of 0.500
Given: 3.79 L of 1.00
nitrate. Calculate the mass of precipitate produced.
For each of the following find:
1. the limiting reactant
2. the amount of product formed in moles and grams
3. the amount of excess in moles and grams
D. 15.0 g of iron (III) oxide reacts with 7.00 g of carbon monoxide.
Find the amount of Iron formed.
Fe2O3(s) + 3CO(g)
2 Fe(s) + 3CO2(g)
E. 7.5 g of sodium chloride reacts with 9.8 g of Lead (II) nitrate.
Find the amount of Lead Chloride formed.
2NaCl(aq) + Pb(NO3)2(aq)
PbCl2(s) + 2NaNO3(aq)
F. 57 g of magnesium hydroxide is added to 86 g of hydrogen phosphate.
Find the amount of Magnesium phosphate formed.
3Mg(OH)2(aq) + 2H3PO4(aq)
Mg3(P04)2(s) + 6H2 O(l)
```