Question

Given,\[{H_2}{O_2}\] turns blackened lead paintings to white colour. In this reaction it oxidizes $PbS$ to $PbS{O_4}$. The number of moles of ${H_2}{O_2}$ needed to oxidize $0.1$moles of $PbS$ is:
A.\[\;1\;\;mole\]
B.$0.1mole$
C.$0.5mole$
D.$0.4mole$

Answer 1

Hint:The number of moles of ${H_2}{O_2}$ needs to be determined from the balanced equation. It is most necessary to write the balanced form of the equation.

Complete step by step answer:
The balanced equation is the form of the equation where the minimum number of reactants and products are shown in such a way that the number of atoms before the reaction and the number of atoms after the reaction remain the same. We need to balance an equation to determine the amount of each reactant and each product.
There is a term called limiting reagent. Limiting reagent is the reagent which determines the amount of product will be formed. Limiting reagent for a reaction is the reagent which is present in less amount.
We know that one mole of an atom means Avogadro number of atoms. It is $6.023*{10^{23}}$ number of atoms. It is generally denoted by “N”. Now let’s write the balanced form of this equation.
 The balanced equation for this reaction is:
$PbS + 4{H_2}{O_2} \to PbS{O_4} + 4{H_2}O$
Here, we can see that in the balanced equation, $4$moles of ${H_2}{O_2}$ is reacting with $1$ mole of PbS to form $PbS{O_4}$ . That means if $0.1$mole of $PbS$ is reacting, $(0.1*4) = 0.4$moles of ${H_2}{O_2}$ will be needed to react with it.
Hence, option (D) is the correct answer.

Note:
It is necessary to write the balanced equation. Without a balanced equation, the solution cannot be done or it can be wrong. There is a term called limiting reagent. Limiting reagent is the reagent which determines the amount of product will be formed. Limiting reagent for a reaction is the reagent which is present in less amount.