Question #1e60f

1 Answer
Ernest Z. · Media Owl
Sep 15, 2014

The mass of PbSO₄ is 7.58 g.

This is a limiting reactant type of problem.

The steps involved are:

  1. Write the balanced equation.
  2. Identify the limiting reactant.
  3. Use conversion factors to convert moles of limiting reactant → moles of PbSO₄ → mass of PbSO₄

Step 1

The balanced equation is

Na₂SO₄ + Pb(NO₃)₂ → 2NaNO₃ + PbSO₄

Step 2

From Na₂SO₄: #"0.0250 L Na"_2"SO"_4 × ("0.100 mol Na"_2"SO"_4)/("1 L Na"_2"SO"_4) × ("1 mol PbSO"_4)/("1 mol Na"_2"SO"_4)# = 0.002 50 mol PbSO₄

From Pb(NO₃)₂: #"0.0400 L Pb(NO"_3")"_2 × ("0.200 mol Pb(NO"_3")"_2)/("1 L Pb(NO"_3")"_2) × ("1 mol PbSO"_4)/("1 mol Pb(NO"_3")"_2)# =
0.008 00 mol PbSO₄

So Na₂SO₄ is the limiting reactant.

Step 3

Mass of PbSO₄ = #"0.002 50 mol PbSO"_4 × ("303.26 g PbSO"_4)/("1 mol PbSO"_4)# = 7.58 g PbSO₄

Related questions
See all questions in Stoichiometry of Reactions Between Ions in Solutions
Impact of this question
2291 views around the world
You can reuse this answer
Creative Commons License