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 Na2SO4×0.100 mol Na2SO41 L Na2SO4×1 mol PbSO41 mol Na2SO4 = 0.002 50 mol PbSO₄

From Pb(NO₃)₂: 0.0400 L Pb(NO3)2×0.200 mol Pb(NO3)21 L Pb(NO3)2×1 mol PbSO41 mol Pb(NO3)2 =
0.008 00 mol PbSO₄

So Na₂SO₄ is the limiting reactant.

Step 3

Mass of PbSO₄ = 0.002 50 mol PbSO4×303.26 g PbSO41 mol PbSO4 = 7.58 g PbSO₄

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