What is the maximum mass of S8 that can be produced by combining 82.0 g of each reactant?

Question

8SO2+16H2S-->3S8+16H2)

Answer 1

Find the limiting reactant. convert both reactants to the same thing

mass of reactant/ molar mass of reactant X mole ratio X molar mass of product

82.0 g SO2 / 48 g/mol  X ( 3 mol S8 / 8 mol SO2)  X 256 g/mol S8  = 164 g S8

82.0 g H2S / 34 g/mol X ( 3 mol S8 / 16 mol H2S) X 256 g/mol S8    =  116 g S8

dihydrogen sulfide is the LR, so you cannot make more than this reactant will produce.