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 [tex]Fe_2O_3[/tex] = 160 g/mole
First we have to calculate the moles of iron.
[tex]\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles[/tex]
Now we have to calculate the moles of [tex]Fe_2O_3[/tex].
The balanced reaction is,
[tex]Fe_2O_3+3CO\rightarrow 2Fe+3CO_2[/tex]
From the balanced reaction, we conclude that
As, 2 moles of iron obtained from 1 mole of [tex]Fe_2O_3[/tex]
So, 0.276 moles of iron obtained from [tex]\frac{0.276}{2}=0.138[/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]
[tex]\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g[/tex]
Therefore, the amount of [tex]Fe_2O_3[/tex] required are, 22.1 grams.
