How many grams of $Mg$ is required to completely reduce $100 \ mL$ of $0.1 \ M$ $NO_3^{-}$ solution using the following reaction:
$NO_3^{-} + Mg \longrightarrow Mg^{2+} + NH_3$

From the given reaction:
$2 KMnO_4 + 3 H_2 SO_4 + 5 H_2 O_2 \longrightarrow K_2 SO_4 + 2 MnSO_4 + 8 H_2 O + 5 O_2$
Find the normality of $H_2 O_2$ solution,if $20 \ mL$ of it is required to react completely with $16 \ mL$ of $0.02 \ M \ KMnO_4$ solution.
$(Molar \ mass \ of \ KMnO_4 = 158 \ g \ mol^{-1})$

Balance the following redox reactions using the oxidation number and ion-electron method:
$(1) H_2S + Fe^{3+} \to Fe^{2+} + S + H^{+}$
$(2) Cu + NO_3^{-} \to Cu^{2+} + NO_2$
$(3) Sn + NO_3^{-} + H^{+} \to Sn^{2+} + NH_4^{+} + H_2O$
$(4) As + NO_3^{-} + H^{+} \to AsO_4^{3-} + NO_2 + H_2O$

Number of moles of $K_2Cr_2O_7$ reduced by one mole of $Sn^{2+}$ ions is

$KMnO_4$ acts as an oxidising agent in acidic medium. The number of moles of $KMnO_4$ that will be needed to react with one mole of sulphide ion $(S^{2-})$ in acidic solution is . . . . . . (in $/5$)?

Moles of $K_2Cr_2O_7$ used to oxidise $1$ $mole$ of $Fe_{0.92}O$ to $Fe^{3+}$ are -