Question

Potassium superoxide $K{{O}_{2}}$ is utilized in the closed system breathing apparatus. Exhaled air contains $C{{O}_{2}}\,and\,{{H}_{2}}O$ both of them are removed and the removal of water molecules generates oxygen for breathing as shown by the reaction $4K{{O}_{2}}\left( s \right)+2{{H}_{2}}O\to 3{{O}_{2}}\left( g \right)+4KOH\left( s \right)$. The potassium hydroxide removes $C{{O}_{2}}$gas from the apparatus by the reaction: $KOH\left( s \right)+C{{O}_{2}}\left( g \right)\to KHC{{O}_{3}}\left( s \right).$ The mass of $K{{O}_{2}}$ which is generating $48g$ of the oxygen gas is:
A. $142g$
B. $153g$
C.$148g$
D. $150g$

Answer 1

Hint:The question is based on the simple concept of mole, atomic mass and Avogadro's hypothesis. In the question we are given with the reaction which releases oxygen, in this reaction we are given the number of moles which are reacting. From these numbers of moles, we will calculate the mass and will use the basic mole concept and unitary method to further solve it.

Complete step-by-step answer: The given balanced reaction for the production of oxygen is represented as follows:
$4K{{O}_{2}}\left( s \right)+2{{H}_{2}}O\to 3{{O}_{2}}\left( g \right)+4KOH\left( s \right)$.
From the above reaction we can say that $4$ moles of $K{{O}_{2}}$ produces $3$moles of ${{O}_{2}}$ on reacting with oxygen.
Now if we convert these moles into mass.
We know that number of moles = $\dfrac{given\,mass}{molar\,mass}$
Therefore, mass of substance = number of moles $\times $molar mass
Let us firstly calculate the moles for $K{{O}_{2}}$.
We know for $K{{O}_{2}}$, given moles in the question = $4$ moles.
Molar mass of $K{{O}_{2}}$= $71g/mol$
Therefore, using formula mass of substance = number of moles $\times $ molar mass
We get mass of $K{{O}_{2}}$= $71\times 4=284g$
Now let us calculate mass of ${{O}_{2}}$
In question Moles of ${{O}_{2}}$= $3$ moles.
Molar mass of ${{O}_{2}}$ is = $32g/mol$
Therefore, using formula mass of substance = number of moles $\times $ molar mass.
We get mass of ${{O}_{2}}$ = $3\times 32=96g$
We have already discussed that from the reaction we can infer that $4$ moles of $K{{O}_{2}}$ produces $3$moles of ${{O}_{2}}$.
Or we can say that $284g$ of $K{{O}_{2}}$ reacts and produces $96g$ ${{O}_{2}}$.
We can now infer that $96g$${{O}_{2}}$ is produced by = $284g$of $K{{O}_{2}}$.
Therefore, $1g\,of\,{{O}_{2}}$is produced by = $\dfrac{284}{96}g\,of\,K{{O}_{2}}$
So, $48g\,of\,{{O}_{2}}$ is produced by = $\dfrac{284}{96}\times 48g\,of\,K{{O}_{2}}$
$\Rightarrow \dfrac{284}{2}=142g$ of $K{{O}_{2}}$.
Hence, $142g$ of $K{{O}_{2}}$ is required to produce $48g\,of\,{{O}_{2}}$.

Hence, the correct option is option. A.

Note: Molar mass can be calculated by adding the masses of the individual element present in the molecule. Its unit is g/mole. Always remember that one mole of a substance contains mass which is equal to its molar mass. For example, 1 mole of oxygen molecule has 32g of mass. Here, 32g is the molar mass of the compound. Oxygen produced by potassium superoxide is very useful for treating patients in an emergency.