If you start with 227.8 grams of iron and 128 grams of oxygen to produce iron oxide, what is the limiting reagent? you will need

Question

2 years ago

10

to balance the equation first. don't forget that oxygen is diatomic (o 2 ) when finding the molar mass. fe o2 -> fe2o3

1 answer:

soldi70 [24.7K] · Answer 1 · 2024-01-02T09:21:26+03:00

Fe is the limiting reactant.

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is

4Fe(s) + 302(g) - ---> 2Fe2O3(s)

Mass of Fe = 227.8 g

Moles of Fe = 227.8gFe*Imol Fe/55.85g Fe = 4.079mol Fe

Mass of oxygen = 128 g

Moles of O2 = 128g02 * 1molo/32g02 = 4molO2

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = 4.079mol Fe*2molFe2O3/4molFe*159.69g Fe2O3/1 mol Fe2O3 = 325.7gFe2O3

Mass of iron (III) oxide produced from oxygen = 4molO2 *2mol Fe2O3/3molO2*159.69gFe2O3/1mol Fe2O3 = 425.84gFe2O3

Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.

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