Precipitation Reaction Problems
1. A solution contains 0.71 mol of aluminum chloride in a 910 mL solution. Calculate
the molarity of all the ions (Al and Cl) in the solution.
Molarity = moles ÷ L
M = 0.71 mol ÷ 0.91 L
= 0.78 mol/L AlCl
∴ 0.71 M is the molarity.
2. 141 mL of 1.2 M aluminum nitrate is mixed with 85 mL of 0.27 M sodium
carbonate.
a. Write a balanced equation for the reaction.
2Al(NO₃)₃ + 3Na₂CO₃ → Al₂(CO₃)₃ + 6NaNO₃
b. Write ionic equations for the reaction.
2Al ³⁺ + 3CO₃ ²⁻ → Al₂(CO₃)₃
and
Na ¹⁺ + NO₃ ¹⁻ → NaNO₃
c. Give the name of any precipitate that has formed.
, Al₂(CO₃)₃ is the precipitate as it is insoluble.
d. Write the net ionic equation for the reaction.
2Al ³⁺ + 3CO₃ ²⁻ → Al₂(CO₃)₃ (s)
e. Find the number of moles of precipitate that has formed.
Molarity [Al(NO₃)₃] = moles ÷ L
moles [Al(NO₃)₃] = M x L
= 1.2 x 0.141
= 0.1692 mol Al(NO₃)₃
moles [Na₂CO₃] = M x L
= 0.27 x 0.085
= 0.0223 mol Na₂CO₃
2Al(NO₃)₃ + 3Na₂CO₃ → Al₂(CO₃)₃ + 6NaNO₃
0.1692 mol Al(NO₃)₃ ÷ 2 mol Al(NO₃)₃ x 1 mol Al₂(CO₃)₃ = 0.0846
mol Al₂(CO₃)₃
∴ 0.0846 mol of precipitate is formed.
f. Find the mass of precipitate that has formed.
1. A solution contains 0.71 mol of aluminum chloride in a 910 mL solution. Calculate
the molarity of all the ions (Al and Cl) in the solution.
Molarity = moles ÷ L
M = 0.71 mol ÷ 0.91 L
= 0.78 mol/L AlCl
∴ 0.71 M is the molarity.
2. 141 mL of 1.2 M aluminum nitrate is mixed with 85 mL of 0.27 M sodium
carbonate.
a. Write a balanced equation for the reaction.
2Al(NO₃)₃ + 3Na₂CO₃ → Al₂(CO₃)₃ + 6NaNO₃
b. Write ionic equations for the reaction.
2Al ³⁺ + 3CO₃ ²⁻ → Al₂(CO₃)₃
and
Na ¹⁺ + NO₃ ¹⁻ → NaNO₃
c. Give the name of any precipitate that has formed.
, Al₂(CO₃)₃ is the precipitate as it is insoluble.
d. Write the net ionic equation for the reaction.
2Al ³⁺ + 3CO₃ ²⁻ → Al₂(CO₃)₃ (s)
e. Find the number of moles of precipitate that has formed.
Molarity [Al(NO₃)₃] = moles ÷ L
moles [Al(NO₃)₃] = M x L
= 1.2 x 0.141
= 0.1692 mol Al(NO₃)₃
moles [Na₂CO₃] = M x L
= 0.27 x 0.085
= 0.0223 mol Na₂CO₃
2Al(NO₃)₃ + 3Na₂CO₃ → Al₂(CO₃)₃ + 6NaNO₃
0.1692 mol Al(NO₃)₃ ÷ 2 mol Al(NO₃)₃ x 1 mol Al₂(CO₃)₃ = 0.0846
mol Al₂(CO₃)₃
∴ 0.0846 mol of precipitate is formed.
f. Find the mass of precipitate that has formed.