Question

$1$ mole of $C{u_2}S$ reduces how many moles of $KMn{O_4}$ ? If the redox reaction is:
$C{u_2}S + KMn{O_4} + {H_2}S{O_4} \to CuS{O_4} + MnS{O_4} + {K_2}S{O_4} + {H_2}O$

Answer 1

Hint : In the above given question we are required to first balance the equation of both the sides of the equation. When we would have balanced the equation then we would have the stoichiometric coefficient which will tell how many moles of $C{u_2}S$ react with $KMn{O_4}$ .

Complete Step By Step Answer:
In the above given question, we are been asked how many moles of $KMn{O_4}$ reacts with one mole of $C{u_2}S$ according to the reaction
$C{u_2}S + KMn{O_4} + {H_2}S{O_4} \to CuS{O_4} + MnS{O_4} + {K_2}S{O_4} + {H_2}O$
Now first of all we would need to balance the chemical reaction
Now the number of atoms of copper and sulphur on the left-hand side are two, eight atoms of oxygen, one atom of manganese and potassium and two atoms of hydrogen.
While on the right-hand side we have one atom of copper, manganese, two atoms of hydrogen, three atoms of sulphur and thirteen atoms of oxygen.
Now both the sides of the equation need to have the same number of atoms.
So, for balancing the equation we will be giving the compounds on both the sides of the reaction stoichiometric coefficient so as to balance the reaction.
We will multiply potassium permanganate with two, while sulphuric acid with four while copper sulphate with two and manganese sulphate with two and multiply water with four.
The reaction after making the given changes will give the reaction as follows: -
$C{u_2}S + 2KMn{O_4} + 4{H_2}S{O_4} \to 2CuS{O_4} + 2MnS{O_4} + {K_2}S{O_4} + 4{H_2}O$
By the above given balanced reaction we can say that one mole of $C{u_2}S$ reacts with two moles of $KMn{O_4}$ .

Note :
Let's understand what a redox reaction is. Redox reaction is defined as the reaction which undergoes both of the reaction reduction, that is gain of electrons and oxidation, that is loss of electrons in a single reaction.