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 

500bc indiais the answer.

All the elements are arranged by melting points on the periodic table. in row they go from metals to gasses. in the row from sodium to argon, sodium is more reactive than the one on its right, magnesium.