Question

Ferric sulphate on heating gives sulphur trioxide. The ratio between the weights of oxygen and sulphur present in $S{O}_3$ obtained by heating $1$ kg of ferric sulphate is:

• A. $2:3$
• B. $1:3$
• C. $3:1$
• D. $3:2$

Answer 1

Answer: (D)

$3:2$

We have

$F{e}_2(S{O}_4{)}_3\to F{e}_2{O}_3+3S{O}_3$

So, 400g of ferric sulphate is producing 240 g of $S{O}_3$

so, 1 kg ( 1000g) of ferric sulphate produce $S{O}_3=\frac{240}{400}\times 1000=600g$

In 80 g of $S{O}_3$, mass of oxygen $=48g$

Mass of sulphur $=32g$,

So, mass of oxygen in 600g of $S{O}_3=\frac{48}{80}\times 600=360g$

mass of sulphur in 600g of $S{O}_3=\frac{32}{80}\times 600=240g$

$\frac{massofoxygen}{massofsulphure}=\frac{360}{240}=\frac{3}{2}$