Question 722d3

Oct 5, 2016

Fe_2O_3(s)+3CO(g)→2Fe(s)+3CO_2(g)#

Considering atomic masses as

$F e \to 56 \text{ g/mol}$

$O \to 16 \text{ g/mol}$

$C \to 12 \text{ g/mol}$

Molar masses of reactants

$F {e}_{2} {O}_{3} \to 2 \cdot 56 + 3 \cdot 16 = 160 \text{ g/mol}$

$C O \to 12 + 16 = 28 \text{ g/mol}$

As per the given equation

1 mol $F {e}_{2} {O}_{3}$ completely reacts with 3 mol $C O$ or in other words

160g $F {e}_{2} {O}_{3}$ completely reacts with $3 \times 28 = 84 g C O$

22.05g $F {e}_{2} {O}_{3}$ completely reacts with $= \frac{84 \times 22.05}{160} = 11.58 g C O$

But the given amount of $C O = 15.56 g$

So the excess amount of $C O$ left after completion reaction is $\left(15.56 - 11.58\right) g = 3.98 g$