How much HCO2H is obtained by reacting 3.97 g of FeS, 1.98 g of H2S, and 1.98 g of CO2 if the reaction results in a 51.0% yield?

Answer 1

3 possible limiting reactants for the equation H₂S +2FeS + CO₂ ⇒ HCO₂H + Fe₂S₃ ( + S ) (possibly)

Since the need for H₂S and CO₂ is 1:1 in moles and we know that CO₂ weighs more, the CO₂ is limiting given equal mass in the feed. The Fe in the FeS is oxidized and may be required or not (as elemental S could be formed as well, in which case the FeS may not be limiting if it is all turned) Oddly enough, it is the LR for the reaction above if it is required as stated. Also, its given amount is just about necessary for 51% yield from the CO₂

I'm going to assume the CO₂ is limiting:

1.98 g CO₂ (1 mole/44.01 g CO₂)( 1 mole HCO₂H/1 mole CO₂)(46.026 g/mole HCO₂H)(51 g act/100 g theory) = g of HCO₂H

If the FeS is Limiting, the calculation is similar.

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