Amount of Ferrous oxalate that reacts with one mole of acidified $KMn{O}_4$is___?

Writing the half reactions of reduction and oxidation,

$\left[2KMn{O}_4+3H_{2}S{O}_4+5e^{-}\to K_{2}S{O}_4+2MnS{O}_4+3H_{2}O\right]\times 3$

$\left[2Fe{C}_2{O}_4+3H_{2}S{O}_4\to F{e}_2{\left(S{O}_4\right)}_3+3H_{2}O+4C{O}_2+3e^{-}\right]\times 5$

On adding the above two half reactions, we get the complete balanced redox reaction.

$6KMn{O}_4+24H_{2}S{O}_4+10Fe{C}_2{O}_4\to 5F{e}_2{\left(S{O}_4\right)}_3+3K_{2}S{O}_4+6MnS{O}_4+24H_{2}O+20C{O}_2$

Then number of moles of Ferrous oxalate that react with one mole of  $KMn{O}_4$is 10/6=1.66

Gram molecular weight of Ferrous oxalate is = 144 gm per mole

So, mass of Ferrous oxalate that react with one mole of $KMn{O}_4$$=1.66\times 144=239gram$