$A$ $20.0 \,mL$ solution containing $0.2 \,g$ impure $H_2O_2$ reacts completely with $0.316 \,g$ of $KMnO_4$ in acidic solution. The purity of $H_2O_2$ (in $\%$) is........... (mol. wt. of $H_2O_2 = 34$; mol. wt. of $KMnO_4 = 158$)

$KMnO_4 + HCl \rightarrow MnCl_2 + Cl_2 + KCl + H_2O$. In the above reaction,how many moles of $H_2O$ would be formed for each mole of $Cl_2$ liberated?

Number of moles of $MnO_4^-$ required to oxidize one mole of ferrous oxalate completely in acidic medium will be .......... $moles$.

For the reaction:
$I^{-} + ClO_{3}^{-} + H_{2}SO_{4} \longrightarrow Cl^{-} + HSO_{4}^{-} + I_{2}$
The correct statement$(s)$ in the balanced equation is/are:
$A$. Stoichiometric coefficient of $HSO_{4}^{-}$ is $6$.
$B$. Iodide is oxidized.
$C$. Sulphur is reduced.
$D$. $H_{2}O$ is one of the products.

In the chemical reaction between stoichiometric quantities of $KMnO_4$ and $KI$ in a weakly basic or neutral solution,what is the number of moles of $I_2$ released for $4$ moles of $KMnO_4$ consumed?

What is the value of $x$ in order to balance the following redox reaction by the ion-electron method?
$x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$