What mass of PbSO4 is produced when 30.0 ml of .1020 M Al2(SO4)3 solution is added to .2442 M Pb(NO3)2?

Question

What mass of PbSO4 is produced when 30.0 ml of .1020 M Al2(SO4)3 solution is added to .2442 M Pb(NO3)2?

Answer 1

First calculate moles of aluminum sulfate. The amount of Pb(NO3)2 doesn't matter since no amount (volume) is specified. You must assume that all of the Al2(SO4)3 reacts and that it is the limiting reagent.

moles Al2(SO4)3 = M Al2(SO4)3 x L Al2(SO4)3 = (0.1020)(0.0300) = 0.00306 moles Al2(SO4)3

Al2(SO4)3 + 3Pb(NO3)2 => 3PbSO4 + 2Al(NO3)3

From the balanced equation we see that there are two moles of PbSO4 for every one mole of Al2(SO4)3.

0.00306 moles Al2(SO4)3 x (2 moles PbSO4 / 1 mole Al2(SO4)3) x (303.3 g PbSO4 / 1 mole PbSO4) = 1.86 g PbSO4