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
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.
370g/[(55.85 x 2) + (13 x 3)] = 2.32 mol
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.
Your calculations are correct, and you obtained the correct answer of 195 grams of CO that can react with 0.370 kg of Fe2O3. Let's break down the steps to understand how you arrived at the answer:
1. First, you converted the given mass of Fe2O3 from kilograms to grams by multiplying it by 1000, since there are 1000 grams in 1 kilogram.
- 0.370 kg x 1000 = 370 g of Fe2O3
2. To calculate the number of moles of Fe2O3, you used the molar mass of Fe2O3, which is the sum of the atomic masses of its constituent elements (2 x atomic mass of Fe + 3 x atomic mass of O).
- Molar mass of Fe2O3 = (55.85 x 2) + (16 x 3) = 159.7 g/mol
- Moles of Fe2O3 = 370 g / 159.7 g/mol = 2.32 mol
3. According to the balanced equation, the stoichiometric ratio between Fe2O3 and CO is 1:3, meaning for every 1 mole of Fe2O3, 3 moles of CO are required.
- Moles of CO = 2.32 mol x 3 = 6.96 mol
4. Finally, to convert the moles of CO to grams, you used the molar mass of CO, which is the sum of the atomic masses of carbon (C) and oxygen (O).
- Molar mass of CO = 12.01 g/mol (atomic mass of C) + 16.00 g/mol (atomic mass of O) = 28.01 g/mol
- Grams of CO = 6.96 mol x 28.01 g/mol = 194.7276 g, which can be rounded to 195 g
Therefore, your calculations are accurate, and you correctly determined that 195 grams of CO can react with 0.370 kg of Fe2O3.