We must compare the two values using the molar ratios in the equation:
0.899 moles Ba(NO3)2 x (1 mole of Na2S04 / 1 mole of Ba(NO3)2) = 0.899 moles of Na2S04
Comparing this value to the value of 0.473 moles of Na2S04 that we calculated in the first part of this problem allows us to determine that we have fewer moles of Na2S04 (when compared to Ba(NO3)2). Therefore, the limiting reactant is Na2S04.
Ba(NO3)2 is in excess and the amount of excess reactant remaining is:
0.899 moles - 0.473 moles = 0.426 moles of excess Ba(NO3)2
0.426 moles Ba(NO3)2 x (235 g/mol) = 100.11 grams of excess Ba(NO3)2
To find the amount of product, we multiply the amount of limiting reactant by the mole ratio of the product/reactant:
0.473 moles Na2S04 x (1 mole BaSO4 / 1 mole Na2S04) = 0.473 moles of BaSO4
mass BaSO4 = moles BaSO4 x molecular weight BaSO4 = 0.473 mol x 233.39 g/mol = 110.39 g
Hope this helps...good luck!