$MnO_4^{-}$ reacts with $A^{x+}$ to form $AO_3^{-}$,$Mn^{2+}$ and $O_2$. One mole of $MnO_4^{-}$ oxidizes $1.25$ moles of $A^{x+}$ to $AO_3^{-}$. What is the value of $x$?

For the following balanced equation,$x, y$ and $z$ will be respectively:
$xMnO_4^- + yH^+ + ze^- \to 2Mn^{2+} + 4H_2O$

Observe the following reaction: $I_2 + 10 HNO_3 \rightarrow 2 HIO_3 + 10 NO_2 + 4 H_2O$. The equivalent weight of $HNO_3$ is (molar mass of $HNO_3 = M$).

Permanganate $(VII)$ ion,$MnO_4^-$,in basic solution oxidises iodide ion,$I^-$,to produce molecular iodine $(I_2)$ and manganese $(IV)$ oxide $(MnO_2)$. Write a balanced ionic equation to represent this redox reaction.

In the balanced redox reaction $MnO_4^- + xFe^{2+} + yH^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$,what are the values of $x$ and $y$?

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$S + HNO_3 \, (Dil.) \longrightarrow H_2SO_4 + NO \uparrow $