Chemistry Question

Question

If you have a solution containing 42.36 g of lead(II) chlorate that is allowed to react completely with a solution containing 7.23 g of sodium sulfate, how many grams of solid precipitate will be formed?

What's the mass of the excess reactant?

Then assuming total precipitation how many moles of each ion remain in the solution?

Answer 1

Lead(II) chlorate = Pb(ClO₃)₂; molar mass 374 g/mol

Sodium sulfate = Na₂SO₄; molar mass 142 g/mol

Lead sulfate = PbSO₄; molar mass = 303 g/mol

Write the correctly balanced equation:

Pb(ClO₃)₂ + Na₂SO₄ ==> 2NaClO₃ + PbSO₄(s)

Find limiting reactant:

Easy way is to divide mols of each reactant by coefficient in balanced equation & whichever value is less

represents the limiting reactant.

For Pb(ClO₃)₂: 42.36 g x 1 mol/374 g = 0.113 mols (÷1->0.113)

For Na₂SO₄: 7.23 g x 1 mol/ 142g = 0.0509 mols (÷1->0.0509) ... LIMITING

Use mols of limiting reactant to find mols & grams of precipitate:

0.0509 mol Na₂SO₄ x 1 mol PbSO₄ / mol Na₂SO₄ = 0.0509 mols PbSO₄ formed

0.0509 mol PbSO₄ x 303 g/mol = 15.4 g PbSO₄ formed

Mass of Pb(ClO₃)₂ excess reactant:

0.0509 mol Na2SO4 x 1 mol Pb(ClO₃)₂ / mol Na2SO4 = 0.0509 mols Pb(ClO₃)₂ used up

0.113 mol - 0.0509 mol = 0.0621 mol left over

0.0621 mols x 374 g/mol = 23.2 g Pb(ClO₃)₂ left over

Moles remaining after complete precipitation: All SO₄^2- will be precipitated. Some of the Pb²⁺ will also be precipitated. All other ions are spectator ions and will not be removed from solution.

.........Initial moles...........Moles used...........Moles remaining

Pb²⁺............0.113................0.0509...................0.0621

SO₄²⁺........0.0509................0.0509..................0

Na⁺ ............0.102...................0.........................0.102

ClO₃^-..........0.113...................0...........................0.113