Steps to solving complex stoichiometry problems – used in a sample problem
Problem: 125 mL of 0.250 M aluminum nitrate and 125 mL of 0.350 M potassium hydroxide
are mixed. Calculate the amount of precipitate formed. Calculate the concentration of all the
ions present after the reaction occurs.
1. Predict products of chemical reaction.
Al(NO3)3 (aq) + KOH (aq)  Al(OH)3 + KNO3
2. Balance equation.
Al(NO3)3 (aq) + 3 KOH (aq)  Al(OH)3 + 3 KNO3
3. Determine the precipitate (consult solubility rules) or find water.
Al(NO3)3 (aq) + 3 KOH (aq)  Al(OH)3 (s) + 3 KNO3 (aq)
4. Calculate the moles of reactants. V x M = moles
Al(NO3)3 (aq) + 3 KOH (aq)  Al(OH)3 (s) + 3 KNO3 (aq)
.125 L
.125 L
x .250 M
x .350 M
0.0313 mol
0.0438 mol
5. Determine the limiting reagent.
1 𝑚𝑜𝑙 𝐴𝑙(𝑂𝐻)3
0.0313 mol 𝐴𝑙(𝑁𝑂3)3 × 1 𝑚𝑜𝑙 𝐴𝑙(𝑁𝑂3)3 = 0.0313 mol Al(OH)3
0.0438 𝑚𝑜𝑙 𝐾𝑂𝐻 ×
1 𝑚𝑜𝑙 𝐴𝑙(𝑂𝐻)3
3 𝑚𝑜𝑙 𝐾𝑂𝐻
= 0.0146 mol Al(OH)3
Reactant that produces smallest amount of product formed is the limiting reagent, so KOH is the
limiting reagent.
6. Find mass of product as directed.
0.0146 𝑚𝑜𝑙 𝐴𝑙(𝑂𝐻)3 ×
78.0037 𝑔
1 𝑚𝑜𝑙
= 1.14 g Al(OH)3 formed
7. Write complete ionic equation.
Al3+ (aq) + 3 NO3- (aq) + 3 K+ (aq) + 3 OH- (aq)  Al(OH)3 (s) + 3 K+ (aq) + 3 NO3- (aq)
8. Cross off spectator ions.
Al3+ (aq) + 3 NO3- (aq) + 3 K+ (aq) + 3 OH- (aq)  Al(OH)3 (s) + 3 K+ (aq) + 3 NO3- (aq)
9. Write net ionic equation.
Al3+ (aq) + 3 OH- (aq)  Al(OH)3 (s)
10. Determine starting number of moles of ions (use #s from step 4 – be sure to multiply if not
1:1).
0.0313 mol Al(NO3)3 contains 0.0313 moles of Al3+ and 0.0939 moles (3 x 0.0313) of NO30.0438 mol KOH contains 0.0438 mol of K+ and 0.0438 mol of OH-
11. Use stoichiometry to determine the moles used (starting with LR from step 5)
Choose ion in limiting reagent that appears in net ionic eq as well.
1 𝑚𝑜𝑙 𝐴𝑙3+
0.0438 mol of OH- × 3 𝑚𝑜𝑙 𝑂𝐻− = 0.0146 mol Al3+ used in reaction
12. Start – use = end number of moles (step 10 – step 11)
Al3+ (aq) +
S
0.0313 mol
U -0.0146 mol
E
0.0167 mol
3 OH- (aq)  Al(OH)3 (s)
0.0438 mol
- 0.0438 mol
0 mol
13. Add the volumes of reactants to determine final volume.
.125 L + .125 L = .250 L
14. Find the concentration of the ion in excess (step 12 / step 13)
[Al3+] = 0.0167 mol / .250 L = 0.0668 M
15. Determine the concentration of spectator ions. (use #s from steps 10 & 13)
The spectator ions are not consumed at all in the reaction. The moles that we start with is the
moles that we end with.
[NO3-] = 0.0939 mol / .250 L = .376 M
[K+] = 0.0438 mol / .250 L = .175 M
16. The concentration of LR ion = 0
[OH-] = 0 M
Now you try!
150 mL of .250 M magnesium sulfate is reacted with 250 mL of 0.400 M silver nitrate.
Calculate the amount of precipitate formed. Calculate the concentration of all the ions present
after the reaction occurs.