How many moles of Fe3O4 are required to supply enough iron to prepare 0.442 mol Fe2O3?
write the balanced equation.
2Fe3O4 + 2H2 >> 3Fe2O3 +2H2O
so to make .442 mol of Iron(III) oxide, you need 2/3 times that.
.137
Well, I'm not sure about the exact number of moles, but I do know that these moles are quite the social creatures. They love to hang out together and make compounds like Fe2O3 and Fe3O4. It's like a mole party, where the iron atoms get busy and mingle. So, if you want to prepare 0.442 mol of Fe2O3, you'll need the appropriate number of Fe3O4 moles to keep the party going. Just remember to invite Avogadro, the king of moles, to this shindig. He'll help you out with all the calculations.
To determine the number of moles of Fe3O4 required, we need to use the balanced chemical equation and the stoichiometry between Fe3O4 and Fe2O3.
The balanced chemical equation for the reaction between Fe3O4 and Fe2O3 is:
3 Fe3O4 + 2 Fe2O3 -> 10 FeO
According to the equation, 3 moles of Fe3O4 react with 2 moles of Fe2O3 to produce 10 moles of FeO.
We are given that we need 0.442 moles of Fe2O3. To find the number of moles of Fe3O4 required, we can set up a proportion using the stoichiometric ratios:
(0.442 mol Fe2O3) / (2 mol Fe2O3) = (x mol Fe3O4) / (3 mol Fe3O4)
Simplifying the proportion:
0.442 / 2 = x / 3
Solving for x, we find:
x = (0.442 / 2) * 3
x = 0.221 mol Fe3O4
Therefore, we would require 0.221 moles of Fe3O4 to supply enough iron to prepare 0.442 mol of Fe2O3.