Question

When $FeS _{2}$ is burnt in air, it converts to $Fe _{2} O _{3}$. What happens to the percentage by weight of iron in the process? $( Fe =56)$

Answer 1

SOLUTION : Calculation of $\%$ of $Fe$ in $FeS _{2}$

Mol. wt. of $FeS _{2}=56+32 \times 2=120 g /$ mole

Since $120 g FeS F _{2}$ contain $5 F + Fe$

$\therefore \quad 100 g Fe S _{2}$ will contain $\frac{56}{120} g =46 \cdot 67 g$

Initial percentage of iron in $FeS _{2}=46 \cdot 67 \%$.

When $FeS _{2}$ is burnt in air, it undergoes oxidation to form $Fe _{2} O _{3}$ and $SO _{2}$.

$4 FeS _{2}+11 O _{2}=2 Fe _{2} O _{3}+8 SO _{2}$

Calculation of $\%$ of $Fe$ in $Fe _{2} O _{3}$

Mol. wt. of $Fe _{2} O _{3}=56 \times 2+16 \times 3=160 g / mole$

$\begin{array}{l}\because \quad 160 g Fe _{2} O _{3} \text { contain } 112 g Fe \\\therefore \quad 100 g Fe Fe _{3} \text { will contain } \frac{112 \times 100}{160} g =70 g .\end{array}$

As a rcsult of reaction there is an increase in the percentage of iron from $46 \cdot 67$ to $70 \cdot 0$