Chemical reaction

Question

If a solution containing 20 g of lead(II) acetate is allowed to react completely with a solution containing 5 g of sodium sulfate, how many grams of solid precipitate will be formed?

mass of solid precipitate:

How many grams of the reactant in excess will remain after the reaction?

mass of excess reactant:

Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles.

Pb^2

C2H3O2^-

Na^+

SO4^2-

Answer 1

First let’s write the balanced equation:

PbAc₂ (aq) + Na₂SO₄ (aq) ===è PbSO₄ (s) + 2NaAc (aq) where Ac stands for acetate ion.

Now, let’s calculate the number of moles of each reactant to determine the limiting reactant.

PbAc₂:  20 g / 325 g/mole = 0.0615 moles

Na₂SO₄: 5 g / 142 g/mole = 0.0352 moles

As the balanced reaction has a 1:1 mole ratio for the reactants, we can see that Na₂SO₄ is the limiting reactant.

To calculate the mass of PbSO₄ produced, use the following calculation:

303 g / 1 mole PbSO₄ x 1 mole Na₂SO₄ / 142 g x 5 g Na₂SO₄ = 10.7 g PbSO₄

The number of moles of excess PbAc₂ will be (0.0615 – 0.0352) = 0.0263 moles or 0.0263 x 325 g = 8.548 g PbAc₂.

Pb+2 = 0.0263 moles remaining

C2H3O2^- = 2 x 0.0263 = 00526 moles remaining

Na+ = 2 x 0.0352 = 0.0704 moles remaining (all the Na₂SO₄ reacts and there are 2 Na for each Na₂SO₄)

SO₄^2- = 0 ( all SO⁴ is tied up in PbSO₄)

Check math; I rounded the atomic masses to the nearest whole number.