What is the maximum mass of s8 that can be produced by combining 77.0 g of each reactant? 8so2+16h2s⟶3s8+16h2o?

The balanced equation for the reaction is as follows
8SO₂ + 16H₂S ---> 3S₈ + 16H₂O
stoichiometry of SO₂ to H₂S is 8:16
we have been given 77.0 g of each reactant 
the reactants react according to molar ratio in other words according to the stoichiometry 
8 mol of SO₂ reacts with 16 mol of H₂S
number of SO₂ moles present - 77.0 g / 64 g/mol = 1.20 mol 
number of H₂S moles present - 77.0 g / 34 g/mol = 2.26 mol 
we need to find the limiting reactant
limiting reactant is fully used up in the reaction 
amount of product formed depends on amount of limiting reactant present 
if SO₂ is the limiting reactant 
if 8 mol of SO₂ reacts with 16 mol of H₂S
then 1.20 mol of SO₂ reacts with - 16/8 x 1.20 mol = 2.40 mol of H₂S
but only 2.26 mol of H₂S are present
which means H₂S is the limiting reactant 
then amount of S₈ formed depends on H₂S present 
stoichiometry of H₂S to S₈ is 16:3
16 mol of H₂S forms 3 mol of S₈
then 2.26 mol of H₂S forms - 3/16 x 2.26 = 0.424 mol
mass of S₈ formed - 0.424 mol x 256 g/mol = 108.5 g
108.5 g of S₈ is formed 

A.