Question

How many grams of Fe2O3 are formed when 51.3 grams of iron, Fe, react completely with oxygen, O2?

Answer 1

Answer: 73.4 g of $Fe_2O_3$ are formed when 51.3 grams of iron react completely with oxygen.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Fe=\frac{51.3g}{56g/mol}=0.92moles$

The balanced chemical reaction is:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

As $Fe$ is the limiting reagent as it limits the formation of product.

According to stoichiometry :

4 moles of $Fe$ produce = 2 moles of $Fe_2O_3$

Thus 0.92 moles of $Fe$ will produce=$\frac{2}{4}\times 0.92=0.46moles$  of $Fe_2O_3$

Mass of $Fe_2O_3=moles\times {\text {Molar mass}}=0.46moles\times 159.69g/mol=73.4g$

Thus 73.4 g of $Fe_2O_3$ are formed when 51.3 grams of iron react completely with oxygen.