What is the maximum mass of S8 that can be produced by combining 75.0 g of each reactant? 8SO2+16H2S-->3S8+16H2O

Question

I need help with solving this for some homework that I am doing in my chemistry class. Thank you for the help

Answer 1

8SO₂+16H₂S-->3S₈+16H₂O

moles SO₂ used = 75.0 g x 1 mole/64 g = 1.17 moles SO₂

moles H₂S used = 75.0 g x 1 mole/34 g = 2.21 moles H₂S

Limiting reactant:  1.17 moles SO₂/8 = 0.146 and 2.21 moles H₂S/16 = 0.138 so H₂S is LIMITING

Maximum moles of S₈ that can be produced = 2.21 moles H₂S x 3 moles S₈/16 moles H₂S = 0.414 moles

Maximum mass of S₈ that can be produced = 0.414 moles x 256 g/mole = 106 g (3 sig.figs)