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.

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

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 $\frac{0.276}{2}=0.138$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

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

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