Question

For the following reaction, if you have 13.2 g of CO and 42.7g of Fe2O3, which is the limiting reagent with regards to Fe production?

Fe2O3 (s) + 3 CO (g) _______> 2 Fe (s) + 3 CO2 (g)

Answer 1

Answer:
`CO` is the limiting reagent and
`Fe_2O_3` is the excess reagent.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Fe_2O_3=(42.7g)/(159.69g/mol)=0.267moles`

`\text{Moles of} CO=(13.2g)/(28g/mol)=0.471moles`

The given balanced equation is :

`Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)`

According to stoichiometry :

3 moles of
`CO` require = 1 mole of
`Fe_2O_3`

Thus 0.471 moles of
`CO` will require=
`(1)/(3)* 0.471=0.157moles` of
`Fe_2O_3`

As given amount of
`Fe_2O_3` is more than the required amount , it is the excess reagent.Thus
`CO` is the limiting reagent as it limits the formation of product.