Question

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?

Answer 1

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

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For $FeS_2$ :

Given mass of $FeS_2$ = 588 g

Molar mass of $FeS_2$ = 120 g/mol

Putting values in equation 1, we get:

$\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol$

• For $O_2$ :

Given mass of $O_2$ = 352 g

Molar mass of $O_2$ = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol$

The chemical equation for the reaction of $FeS_2$ and oxygen gas follows:

$FeS_2+O_2\rightarrow Fe_2O_3+SO_2$

By Stoichiometry of the reaction:

1 mole of $FeS_2$ reacts with 1 mole of oxygen gas

So, 4.9 moles of $FeS_2$ will react with = $\frac{1}{1}\times 4.9=4.9mol$ of oxygen gas

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

Thus, $FeS_2$ is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of $FeS_2$ produces 1 mole of iron (III) oxide

So, 4.9 moles of $FeS_2$ will produce = $\frac{1}{1}\times 4.9=4.9moles$ 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:

$4.9mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(4.9mol\times 159.7g/mol)=782.5g$

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