Fe3O4 + 4 H2 -> 3 Fe + 4 H2O
What mass (g) of Fe3O4 would be needed to
react with 1.38 x 10^3 mol H2?
Damon answered your question on your first post and I answered the one on moles below that. What's the problem?
The answer doesn't match with Damon's post. His mass came out way too high.
Do you want me to look at it and repost here if needed?
Yes please!
Yes, Damon's response, as noted in that response, is he though it was 1.38E3 L (and not mols).
Here is the question again.
Fe3O4 + 4 H2 -> 3 Fe + 4 H2O
What mass (g) of Fe3O4 would be needed to
react with 1.38 x 10^3 mol H2?
1.38E3 mols H2 x (1 mol Fe3O4/4 mols H2) = 345 mols Fe3O4
g Fe3O4 = mols Fe3O4 x molar mass Fe3O4
g Fe3O4 = 345 x 231.54 = 7.99E4 grams.
I just looked at Damon's response again and I don't see anything wrong with it. The answer I posted is almost the same as the answer he posted. The only difference is he rounded the molar mass of Fe3O4 from 231.54 (which I used) to 232. Other than that there is no difference.
To determine the mass of Fe3O4 needed to react with a given amount of H2, we can use the balanced chemical equation and molar ratios.
The balanced equation tells us that the ratio of Fe3O4 to H2 is 1:4. This means that for every 4 moles of H2, we need 1 mole of Fe3O4.
Given that we have 1.38 x 10^3 mol of H2, we can set up a proportion to find the amount of Fe3O4 required:
(1 mol Fe3O4 / 4 mol H2) = (x mol Fe3O4 / 1.38 x 10^3 mol H2)
To solve for x, we can cross-multiply and divide:
x = (1 mol Fe3O4 / 4 mol H2) * (1.38 x 10^3 mol H2)
x = 1.38 x 10^3 mol H2 * (1 mol Fe3O4 / 4 mol H2)
x = 3.45 x 10^2 mol Fe3O4
The equation shows that 1 mole of Fe3O4 has a molar mass of 3 * (55.845 g/mol) + 4 * (16.00 g/mol) = 231.532 g/mol.
To find the mass of Fe3O4 required, we multiply the number of moles by the molar mass:
mass = 3.45 x 10^2 mol * 231.532 g/mol
mass ≈ 7.98 x 10^4 g
Therefore, approximately 7.98 x 10^4 grams of Fe3O4 would be needed to react with 1.38 x 10^3 mol of H2.