Question

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

• A. $1.2gms$
• B. $2.4gms$
• C. $3.6gms$
• D. none of these

Answer 1

The correct option is A $1.2gms$

The reaction between iron and oxygen is given by

$4Fe+3O₂\to 2Fe₂O₃$

To calculate moles of 2.79 g of iron

Moles = mass/molar mass
$=2.79/55.8$
$=0.05moles$

$MoleratiobetweenFeandFe₂O₃is4:2=2:1$

$thus,themolesoftherust\left(Fe₂O₃\right)=0.05/2=0.025moles$

To calculate mass of $Fe₂O₃$

Mass = moles × molar mass

$=0.025\times (55.8\times 2+16\times 3)$
$=0.025\times 159.6$
$=3.99$

Hence, the mass of the rust formed $\left(Fe₂O₃\right)$ is 3.99 grams

To find the mass of oxygen in the rust $\left(Fe₂O₃\right)$

The % by mass of the oxygen $48/159.6\times 100$ = 30.08%

Hence, the mass of the oxygen in the rust is:

$30.08/100\times 3.99g$ = $1.2g$

Therefore the mass of the oxygen in the rust formed is $1.2g$