It's a limiting reagent problem.
8mol*(64.07 g/mol)=512.56 g
16mol*(34.07 g/mol)=545.12 g
Therefore, SO2 is the limiting reagent. Figure out the rest
What is the maximum mass of S8 that can be produced by combining 85.0 g of each reactant?
8SO2 + 16H2S yields 3S8 +16H2O
2 answers
I don't believe Junday into account the 85 grams of each reactant.
mols SO2 = 85/64 = about about 1.3
mols H2S = 85/34 = about about 2.5
Calculate LR from that and I believe the LR is H2S. Check it out.
mols SO2 = 85/64 = about about 1.3
mols H2S = 85/34 = about about 2.5
Calculate LR from that and I believe the LR is H2S. Check it out.