Question

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

4 Fe + 3 O2 --> 2 Fe2O3

Answer 1

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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Fe=(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=
`(2)/(4)* 0.92=0.46moles` of
`Fe_2O_3`

Mass of
`Fe_2O_3=moles* {\text {Molar mass}}=0.46moles* 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.