) A quantity of 10.0 g Iron is reacted with 20.0 g Oxygen gas. Calculate the the amount of Iron(III) oxide that could be made. 4Fe(s) + 3O2(g)  2Fe2O3(s).

Question

when the reaction is finished, 12.7 g of iron oxide is measured. calculate the percent yield

Answer 1

You asked the first part of this question earlier, and here is the final part of my response.

0.1792 mol Fe x 2 mol Fe₂O₃ / 4 mol Fe = 0.0896 mols Fe₂O₃ (3 sig. figs.)

If you want the mass of Fe₂O₃, then use the molar mass of Fe₂O₃ to convert:

0.0896 mol Fe₂O₃ x 159.7 g / mol = 14.3 g Fe₂O₃ (3 sig. figs.)

So theoretical yield of Fe2O3 is 14.3 g

Percent yield = actual yield / theoretical yield (x100%)

Percent yield = 12.7 g / 14.3 g (x100%)

Percent yield = 88.8% yield