What is the maximum mass of S8 that can be produced by combining 83.0 g of each reactant? 8SO2+16H2S⟶3S8+16H2O

Explanation:

Find number of moles in each reactant

SO2 moles=83/64 mol

H2S moles=83/34 mol

SO2 : H2S

8 : 16

1 : 2

(83/64) mol : (83/64)×2 mol

we need (83/32) moles of H2S to fully react with SO2

no of H2S moles needed to fully react with SO2> no of H2S moles involved in the reaction

83/32 mol> 83/34 mol

so we can see that H2S is the Limiting reagent

we should calculate no of product moles created from the reaction according to

Limiting reagent

H2S : S8

16.:3

1. :3/16

(83/34): 83/34×3/16

Molar mass of S8

=32×8

=256gmol-1

Maximum mass of S8 produced from the reaction;

n=m/M

m=(249/544)×256

=117.17 g