Question

$2.79$ g of iron (atomic weight $=55.8$) is completely converted to rust. The weight of oxygen in the rust is:

• A. $1.2$ g
• B. $2.4$ g
• C. $4.8$ g
• D. $3.6$ g

Answer 1

Answer: (A)

$1.2$ g

The reaction for the formation of rust is $4Fe+3O_2\to 2F{e}_2{O}_3$.

In one molecule of $F{e}_2{O}_3$ (molecular weight $159.6$ g/mol), two iron atoms ($111.6$ g) and three oxygen atoms ($48$ g) are present.

Thus, $111.6$ g of iron corresponds to $48$ g of oxygen.

$1$ g of iron will correspond to $\frac{48}{111.6}$ g of oxygen.

Hence, $2.79$ g of iron will correspond to $\frac{48}{111.6}\times 2.79=1.2$ g of oxygen.