Question

One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5 g of Fe if the reaction occurs in the presence of excess CO?

a.11.1 g

b.22.1 g

c.30.0 g

d.44.2 g

Answer 1

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of
`Fe_2O_3` = 160 g/mole

First we have to calculate the moles of iron.

`\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=(15.5g)/(56g/mole)=0.276moles`

Now we have to calculate the moles of
`Fe_2O_3`.

The balanced reaction is,

`Fe_2O_3+3CO\rightarrow 2Fe+3CO_2`

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of
`Fe_2O_3`

So, 0.276 moles of iron obtained from
`(0.276)/(2)=0.138` mole of
`Fe_2O_3`

Now we have to calculate the mass of
`Fe_2O_3`

`\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3* \text{Molar mass of }Fe_2O_3`

`\text{Mass of }Fe_2O_3=(0.138mole)* (160g/mole)=22.08g=22.1g`

Therefore, the amount of
`Fe_2O_3` required are, 22.1 grams.