To precipitate all of the Pb2+ cations found in an 68 ml solution of 0.796 molarity Pb(NO3)2 you will need to add ________ grams of Na2SO4 .

Question

I honestly have no idea where to even begin when trying to solve it.

Answer 1

You ALWAYS begin by writing down the correctly balanced equation for the reaction taking place:

Pb(NO₃)₂(aq) + Na₂SO₄(aq) ===> PbSO₄(s) + 2NaNO₃(aq) ... balanced equation

Now, you find the moles of Pb(NO₃)₂ that are present:

68 ml = 0.068 L

0.068 L x 0.796 moles/L = 0.0541 moles Pb(NO₃)₂ present

Next, using the balanced equation, we find the moles of Na₂SO₄ needed to completely react with this:

0.0541 moles Pb(NO₃)₂ x 1 mol Na₂SO₄/mol Pb(NO₃)₂ = 0.0541 moles Na₂SO₄ needed

Finally, we convert the moles of Na₂SO₄ to grams of Na₂SO₄:

0.0541 moles Na₂SO₄ x 142.04 g/mol = 7.689 g = 7.69 g Na₂SO₄ (to 3 significant figures)