250 x 0.82 = estimated 205 g starting FeS2. Convert 205 g to mols = about 2 mol but you should confirm that.
Then mols H2SO4 = 2 mols FeS2 x (8 mols SO2/2 mol FeS2) x 0.95 x (2 mol SO3/2 SO2) x 0.90 x (1 mol H2SO4/1 mol SO3) x 0.90= mols H2SO4.
Convert to grams.
How many grams of pure H2SO4 can be obtained from 250 grams of iron ore if the ore is 82% FeS2? The reactions involved are given below:
4FeS2 + 11 O2 --> 2 Fe2O3 + 8 SO2 (95% efficient)
2SO2 + O2 --> 2SO3 (90% efficient)
SO3 + H2O --> H2SO4 (90% efficient)
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