My educated guess is that you didn't put in the correct molar mass Fe2O3. It should be
[(2*55.85 + 3*(16)] = about 159.69 or so but that may be a typo since your answer of 2.32 mols is correct (although I obtained 2.317 and I always carry an extra place and round at the end.
Then 2.317 x 3 x 28.01 = 194.69 g CO which rounds to 195.
Other than the typo I don't see anything wrong.
Please tell me what I did wrong on this problem:
Given the equation Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g),
Calculate the number of grams of CO that can react with 0.370kg of Fe2O3
Here's what I did:
.370kg x 1000 = 370g of Fe2O3
370g/[(55.85 x 2) + (13 x 3)] = 2.32 mol
2.32 mol x 3 = 6.96 mol of CO
6.96 mol x (12.01 + 16) = 195 g of CO
3 answers
370g/[(55.85 x 2) + (13 x 3)] = 2.32 mol
Don't you mean 16 * 3 not 13 * 3 ?
Don't you mean 16 * 3 not 13 * 3 ?
I think that 13 was a typo. The correct molar mass Fe2O3 is 159.69 and
370g/159.60 = 2.317 which rounds to 2.32 and the remaining part of the math is correct.
370g/159.60 = 2.317 which rounds to 2.32 and the remaining part of the math is correct.