Question

One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Suppose that 1.79 × 103 kg of Fe is obtained from a 2.86 × 103 kg sample of Fe2O3. Assuming that the reaction goes to completion, what is the percent purity of Fe2O3 in the original sample?

Answer 1

Answer : The percent purity of
`Fe_2O_3` in the original sample is 87.94 %

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=(1.79* 10^3kg)/(55.8g/mole)=(1.79* 10^3* 1000g)/(55.8g/mole)=3.15* 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,
`3.15* 10^4mole` of Fe produced from
`(3.15* 10^4)/(2)=15750` 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=(15750moles)* (159.69g/mole)=2.515* 10^6g=2.515* 10^3kg`

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

Mass of original sample =
`2.86* 10^3kg`

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

`\text{Percent purity}=(2.515* 10^3kg)/(2.86* 10^3kg)* 100=87.94\%`

Therefore, the percent purity of
`Fe_2O_3` in the original sample is 87.94 %