Solution Stoichiometry

Themes > Science > Chemistry > General Chemistry > Solution and Solubility > Solubility of Solutes and Aqueous Solutions > Solution Index > Solution Stoichiometry

Working stoichiometry problems using solutions isn't much harder than a normal stoichiometry problem. Rather than using the molecular weight to convert a weight to moles, you use the volume and molarity of the solutions to get the moles of the reactants. Once you have that, you can work out the moles of products normally.

Example 1: You add 500 ml of 0.100 M AgNO₃ solution to a solution containing an excess of Cl^- ion. How much AgCl precipitate will you form?

Solution 1: First, work out the number of moles of silver ion from the molarity. For each mole of AgNO₃, you'll form 1 mole of Ag⁺ ion, so the concentration of silver ion is the same as the concentration of the AgNO₃. Convert this to moles by multiplying concentration by volume:

moles Ag⁺ = 0.100 mol/L * 500 ml * 1 L/1000 ml

moles Ag⁺ = 0.0500 moles

Next, write down the net ionic equation for the reaction:

Ag⁺(aq) + Cl^-(aq) -> AgCl(s)

Now we just do the normal stoichiometry problem. The ratio between Ag⁺ and AgCl is 1/1, so

0.0500 moles Ag⁺ * 1 AgCl/1 Ag⁺ = 0.0500 moles AgCl

Example 2: If you mix 200 ml of 0.100 M Pb(NO₃)₂ and 300 ml of 0.200 M MgCl₂, how much PbCl₂ precipitate will you form?

Solution 2: This is a limiting reagent problem, and is worked the same way any other such problem is worked. First, work out how much of each ion there is in solution: for lead it is easy, but remember that the ion concentration of Cl^- from the MgCl₂ is twice that of the MgCl₂ concentration since there are two Cl^- for each MgCl₂ molecule

moles Pb⁺² = 0.100 mol Pb(NO₃)₂/L * 200 ml * 1 L/1000 ml = 0.0200 mol * 1 Pb⁺²/1 Pb(NO₃)₂ = 0.0200 moles Pb⁺²

moles Cl^- = 0.200 mol MgCl₂/L * 300 ml * 1 L/1000 ml * 2 Cl^-/1 MgCl₂ = 0.120 moles Cl^-

Now, work the problem twice, once assuming that the lead runs out first, once assuming that the chloride runs out first. The balanced chemical equation is

Pb⁺²(aq) + 2Cl^-(aq) -> PbCl₂(s)

so we will form

0.0200 moles Pb⁺² * 1 PbCl₂/1 Pb⁺² = 0.0200 moles PbCl₂ if lead runs out first

0.120 moles Cl^- * 1 PbCl₂/2 Cl^- = 0.0600 moles PbCl₂ if chloride runs out first

Since we make less product if the lead runs out first, lead is the limiting reagent and thus we form 0.0200 moles PbCl₂

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