Question

If 44.9 g of iron(III) oxide and 19.9 g of carbon react, what is the theoretical yield in grams of iron metal? 2 Fe2O3(s) + 3 C(s) → 4 Fe (s) + 3 CO2(g)

Answer 1

Approx. `30*g`........

Explanation:

`2Fe_2O_3(s) + 3C(s) + Delta rarr 4Fe + 3CO_2(g)uarr`

And so we work out the stoichiometric quantities of ferric oxide and carbon....

`"Moles of ferric oxide"=(44.9*g)/(159.69*g*mol^-1)=0.281*mol`

`"Moles of carbon"=(19.9*g)/(12.011*g*mol^-1)=1.66*mol`

Now, clearly, the carbon reagent is in excess (which makes sense on the basis that carbon is cheap and plentiful), and the `"ferric oxide"` is the limiting reagent.

And thus, given 100% yield, we should get `2xx0.281*molxx55.85*g*mol^-1=??*g`