Find the theoretical yield (g) of Fe2O3, produced by the rxn of 100.0 g of Fe and 50 g of O2. Actual = 134.08

Find the theoretical yield (g) of Fe2O3, produced by the rxn of 100.0 g of Fe and 50 g of O2. The actual yield is 134.08 g, what is the percent yeild of Fe2O3?

3O2 + 4Fe = 2Fe2O3

Answer:
Let's see here. We have

moles of iron = 100/55.847 = 1.791 moles
moles of oxygen = 50 gm/15.999 = 3.125 moles

We use 6 moles of oxygen for every mole of iron. This reaction will be limited by the amount of iron. If we use 1.791 moles of iron, we will also consume

1.791*6/4 = 2.686 moles of oxygen.

The mass of material used will be

100*gm of iron + 1.343 *15.99 = 142.972 grams of reactants.

The yield is 134.08/142.972 = 93.8%

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