Answer:
Explanation:
2 H₂S(g) +S0₂(g) = 3 S(s) + 2H₂0(g)
2 x 34 g 64 g 3 x 32 g
68 g of H₂S reacts with 64 g of S0₂
3.89 g of H₂S reacts with 64 x 3 .89 / 68 g of S0₂
3.89 g of H₂S reacts with 3.66 g of S0₂
S0₂ given is 4.11 g , so it is in excess .
Hence H₂S is limiting reagent .
68 g of H₂S reacts with S0₂ to give 96 g of Sulphur
3.89 g of H₂S reacts with S0₂ to give 96 x 3.89 / 68 g of Sulphur
3.89 g of H₂S reacts with S0₂ to give 96 x 3.89 / 68 g of Sulphur
5.49 g of Sulphur is produced .
Actual yield is 4.89
percentage yield = 4.89 x 100 / 5.49
= 89 % .