If you have 20 grams of lead (iv) sulfate and 15 grams of lithium nitrate, what is the maximum of lithium sulfate that you can produce

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.