Respuesta :
Answer : The percent purity of [tex]Fe_2O_3[/tex] in the original sample is 87.94 %
Explanation :
The given balanced chemical reaction is:
[tex]Fe_2O3+3CO\rightarrow 2Fe+3CO_2[/tex]
First we have to calculate the mass of Fe.
[tex]\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}[/tex]
Molar mass of Fe = 55.8 g/mole
[tex]\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole[/tex]
Now we have to calculate the moles of [tex]Fe_2O_3[/tex]
From the balanced chemical reaction we conclude that,
As, 2 moles of Fe produced from 1 mole of [tex]Fe_2O_3[/tex]
So, [tex]3.15\times 10^4mole[/tex] of Fe produced from [tex]\frac{3.15\times 10^4}{2}=15750[/tex] mole of [tex]Fe_2O_3[/tex]
Now we have to calculate the mass of [tex]Fe_2O_3[/tex]
[tex]\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3[/tex]
Molar mass of [tex]Fe_2O_3[/tex] = 159.69 g/mole
[tex]\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg[/tex]
Now we have to calculate the percent purity of [tex]Fe_2O_3[/tex] in the original sample.
Mass of original sample = [tex]2.86\times 10^3kg[/tex]
[tex]\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100[/tex]
[tex]\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%[/tex]
Therefore, the percent purity of [tex]Fe_2O_3[/tex] in the original sample is 87.94 %
