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0.10 g of Fe3O4 reacts with excess O2 to give Fe2O3 in 20.36% yield according to...

0.10 g of Fe3O4 reacts with excess O2 to give Fe2O3 in 20.36% yield according to the following balanced equation. Calculate the theoretical yield and the actual yield of Fe2O3(s).

4Fe3O4(s) + O2(g) --> 6Fe2O3(s)

Solutions

Expert Solution

0.10 g of Fe3O4 reacts with excess O2 to give Fe2O3 in 20.36% yield according to the following balanced equation. Calculate the theoretical yield and the actual yield of Fe2O3(s).

4Fe3O4(s) + O2(g) --> 6Fe2O3(s)

Solution:

Molecular weight of Fe3O4 = 231.53 g / mole

Moles of Fe3O4 used = Mass / Molecular weight = 0.10 / 231.53 = 4.32 X 10^-4 moles

As per balanced equation, 4moles of Fe3O4 wll react with O2 to give 6 moles of Fe2O3.

So one mole of Fe3O4 will react to give 6/4 moles of Fe2O3

Therefore 4.32 X 10^-4 moles of Fe3O4 will react with excess oxygen to give = 4.32 X 10^-4 X 6 / 4 moles of Fe2O3

                         = 6.48 X 10^-4 moles of Fe2O3

Molecular weight of Fe2O3 = 159.68 g / mole

Theoretical yield = Moles X molecular weight of Fe2O3 = 0.1034 grams of Fe2O3

Actual yield = 20.36% X 0.1034 grams of Fe2O3 = 0.0211 grams of Fe2O3

queen_honey_blossom answered 2 years ago
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