Question

A 75.0 ml aliquot of 0.100 M aqueous potassium chromate is mixed with 100.0 ml of 0.100 M aqueous barium nitrate. Show set up of problem, with each number labeled with substance and units, as well as the final answer. Write a balance chemical equation. What precipitate is formed? What mass of the precipitate is produced?

Answer 1

Potassium chromate reacts with barium nitrate.

Molecular equation:
K₂CrO₄ (aq) + Ba(NO₃)₂ (aq)    BaCrO₄ (s) + 2 KNO3 (aq)

BaCrO₄ is the precipatate.

Now,

Moles = Molarity x Volume (L)

moles of potassium chromate = 0.100 M x 0.075 L = 0.0075 moles

moles of barium nitrate = 0.100 M x 0.100 L = 0.0100 moles

moles of potassium chromate (= 0.0075 moles) is less and hence the limiting reagent.

Again,

K₂CrO₄ (aq) + Ba(NO₃)₂ (aq)    BaCrO₄ (s) + 2 KNO3 (aq)

In the reaction equation,

1 mole of K₂CrO₄ (aq) reacts with 1 mole of Ba(NO₃)₂ (aq) to produce 1 mole of BaCrO₄ (s).

Or, 1 mole of K₂CrO₄ (aq) produces 1 mole of BaCrO₄ (s).

Or, 0.0075 mole of K₂CrO₄ (aq) produces 0.0075 mole of BaCrO₄ (s).

Molar mass of BaCrO₄ = 194.19 g/mol

So, 1 mole of BaCrO₄ = 194.19 g

0.0075 mole of BaCrO₄ = 0.0075 x 194.19 g
= 1.46 g