Question

What mass of iron(III) oxide is required to produce 12.8g of iron? 2Fe2O3 + 3C → 4Fe + 3CO2

Answer 1

`18.12g` `Fe_2O_3`

Explanation:

Regardless of the other reactant quantities necessary, and the final products, the conservation of mass requires that we have 12.8g of Fe in the iron oxide if we expect to get at least 12.8 g of Fe metal product.

SO, the question is, "What mass of `Fe_2O_3` contains 12.8g Fe?"

By molar ratio, we have 2 moles of Fe per mole of `Fe_2O_3`.

One mole of `Fe_2O_3`is 158grams. The two moles of Fe in comprise 111.6g of it, so we need roughly `1/10` that amount.

EXACTLY, we need `12.8g Fe xx (Fe_2O_3"moles")/(111.6g Fe) xx 158g/"mole" Fe_2O_3`

`= 18.12g` `Fe_2O_3`

CHECK:
Mass Ratio of `Fe:Fe_2O_3 = 111.6/158 = 0.706`

`0.705 xx 18.12g Fe_2O_3 = 12.8g Fe`

Answer 2

Well, we got `2Fe_2O_3(s) + 3C(s) + Deltararr4Fe(l) + 3CO_2(g)`

Explanation:

And as written `320*g` of ferric oxide is reduced by `36*g` of carbon, to give `223.4*g` and `132*g` carbon dioxide.....

As applies to ANY chemical reaction, CHARGE and MASS are absolutely conserved.....

Now (finally), we address your question. We get a molar quantity with respect to iron of `(12.8*g)/(55.85*g*mol^-1)=0.229*mol`.

And given the molar quantity of iron, THERE MUST have been a `(0.229*mol)/2` quantity with respect to ferric oxide.....

i.e. `(0.229*mol)/2xx159.69*g*mol^-1-=18.3*g`.

Do you agree?