Asked by Anonymous

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?
Answered by DrBob222
Damon answered your question on your first post and I answered the one on moles below that. What's the problem?
Answered by Anonymous
The answer doesn't match with Damon's post. His mass came out way too high.
Answered by DrBob222
Do you want me to look at it and repost here if needed?
Answered by Anonymous
Yes please!
Answered by DrBob222
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.
<b>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? </b>

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.
Answered by DrBob222
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.
There are no AI answers yet. The ability to request AI answers is coming soon!