11 grams
The balanced reaction is
Pb(SO4)2 + 4 LiNO3 ==> Pb(NO3)4 + 2 Li2SO4
First, let's calculate the number of moles of each reactant we have, start by calculating their molar masses using the atomic weights of the involved elements:
Atomic weight lead = 207.2
Atomic weight sulfur = 32.065
Atomic weight oxygen = 15.999
Atomic weight lithium = 6.941
Atomic weight nitrogen = 14.0067
Molar mass Pb(SO4)2 = 207.2 + (32.065 + 15.999*4)*2 = 399.322 g/mol
Molar mass LiNO3 = 6.941 + 14.0067 + 15.999*3 = 68.9447 g/mol
Molar mass Li2SO4 = 6.941*2 + 32.065 + 15.999*4 = 109.943 g/mol
Moles Pb(SO4)2 = 20 g / 399.322 g/mol = 0.050084894 mol
Moles LiNO3 = 15 g / 68.9447 g/mol = 0.217565672 mol
Looking at the balanced reaction, for every mole of Pb(SO4)2 used, we need 4 moles of LiNO3. Since we have 0.050084894 mol of Pb(SO4)2, we need 0.050084894 mol * 4 = 0.200339576 mol of LiNO3 which is less than the available quantity of 0.217565672 mol. So the limiting reactant is Pb(SO4)2.
Now looking at the balanced reaction, for every mole of Pb(SO4)2 used, we produce 2 moles of Li2SO4. So we'll have 0.050084894 mol * 2 = 0.100169788 mol of product, which will mass 0.100169788 mol * 109.943 g/mol = 11.01296698 g. Rounding to 2 significant digits gives us 11 grams of Lithium sulfate.
