Question

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is Fe2O3 1 3CO ¡ 2Fe 1 3CO2 Suppose that 1.64 3 103 kg of Fe are obtained from a 2.62 3 103-kg sample of Fe 2 O 3 . Assuming that the reaction goes to completion, what is the percent purity of Fe 2 O 3 in the original sample?

Answer 1

Answer : The percent purity of
`Fe_2O_3` in the original sample is 89.6 %

Explanation :

The given balanced chemical reaction is:

`Fe_2O3+3CO\rightarrow 2Fe+3CO_2`

First we have to calculate the mass of Fe.

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

Molar mass of Fe = 55.8 g/mole

`\text{Moles of }Fe=(1.64* 10^3kg)/(55.8g/mole)=(1.64* 10^3* 1000g)/(55.8g/mole)=2.94* 10^4mole`

Now we have to calculate the moles of
`Fe_2O_3`

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of
`Fe_2O_3`

So,
`2.94* 10^4mole` of Fe produced from
`(2.94* 10^4)/(2)=14700` 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`

Molar mass of
`Fe_2O_3` = 159.69 g/mole

`\text{ Mass of }Fe_2O_3=(14700moles)* (159.69g/mole)=2.347* 10^6g=2.347* 10^3kg`

Now we have to calculate the percent purity of
`Fe_2O_3` in the original sample.

Mass of original sample =
`2.62* 10^3kg`

`\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}* 100`

`\text{Percent purity}=(2.347* 10^3kg)/(2.62* 10^3kg)* 100=89.6\%`

Therefore, the percent purity of
`Fe_2O_3` in the original sample is 89.6 %