Question

Im really stuck on how to solve these problems

What is the maximum mass of S8 that can be produced by combining 83.0 g of each reactant? 850, + 16H,S → 3S8 + 16H20 Number g S

Answer 1

No of moles of SO_2 = 83.0 g times 1 mol/64.07 g = 1.30 mol No of moles of H_2 S = 83.0 g times 1 mol/34.1 g = 2.43 mol Equation: 8 SO_2 + 16 H_2 S 3S_8 + 16 H_2 O No of moles of S_8 produced from SO_2 = 1.30 mol SO_2 times 3 mol S_8/8 mol SO_2 = 0.488 mol H_2 SO_4 No of moles of S_8 produced from H_2 S = 2.43 mol H_2 S times 3 mol S_8/16 mol H_2 S = 0.456 mol S_8 Since H_2S limits the production of S_8, H_2 S acts as the limiting reactant No of moles of S_8 produced = 0.456 mol S_8 Theoretical yield of S_8 = 0.456 mol S_8 times 256.6 g/1 mol S_8