Question

If the 1.58g of $KMn{O}_4$ in acidic medium completely reacts with ferrous oxalate what weight (in gm) of ferrous oxalate is required?

• A. 2.73 gm
• B. 4.73 gm
• C. 11.19 gm
• D. 5.62 gm

Answer 1

The correct option is A 2.73 gm

$Fe{C}_2{O}_4\Rightarrow n_f=3$

$KMn{O}_4\Rightarrow n_f=5$

Moles of $KMn{O}_4=\frac{1.58}{158}=0.01$

Equivalents of $KMn{O}_4=moles\times n_f=0.01\times 5=0.05$

Equivalents of $Fe{C}_2{O}_4=moles\times n_f=n\times 3=0.05$

$\Rightarrow n=\frac{0.05}{3}$

Weight of $Fe{C}_2{O}_4=n\times 142=2.73gm$