If 588 grams of FeS2 is allowed to react with 352 grams of O2 according to the following equation, how many grams of Fe2O3 are produced?

The mass of iron (III) oxide  produced is 782.5 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of = 588 g

Molar mass of = 120 g/mol

Putting values in equation 1, we get:

Given mass of = 352 g

Molar mass of = 32 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of and oxygen gas follows:

By Stoichiometry of the reaction:

1 mole of reacts with 1 mole of oxygen gas

So, 4.9 moles of will react with = of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of produces 1 mole of iron (III) oxide

So, 4.9 moles of will produce = of iron (III) oxide

Now, calculating the mass of iron (III) oxide  from equation 1, we get:

Molar mass of iron (III) oxide  = 159.7 g/mol

Moles of iron (III) oxide  = 4.9 moles

Putting values in equation 1, we get:

Hence, the mass of iron (III) oxide  produced is 782.5 grams