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.