Question

Potassium hydroxide $(K{{O}_{2}})$ is used in rebreathing gas masks to generate oxygen:
\[K{{O}_{2}}(s)+{{H}_{2}}O(l)\to KOH(s)+{{O}_{2}}(g)\]
If a reaction vessel contains 0.158 mol of $(K{{O}_{2}})$ and the 0.10 mol of ${{H}_{2}}O$. How many moles of ${{O}_{2}}$ can be produced?
A.0.1185
B.0.1125
C.0.1265
D.0.2008

Answer 1

Hint:To solve this type of question it is important to first write the balanced chemical reaction involved in the given process. After writing the balanced chemical equation it will be easy to calculate the number of moles.

Complete answer:
Limiting reagent is defined as the substance which are entirely consumed in the completion of the chemical reaction. They are also known as limiting reactants and limiting agents.
The balanced chemical reaction involved is mentioned below:
\[4K{{O}_{2}}+2{{H}_{2}}O\to 4KOH+3{{O}_{2}}\]
Given the number of moles of $(K{{O}_{2}})$ is 0.158 and the given number of moles of water is 0.10.
The ratio of moles and the stoichiometry coefficient of $(K{{O}_{2}})$and water will be $\dfrac{0.158}{4}\text{ and }\dfrac{0.10}{2}=0.0395\text{ and 0}\text{.05}$ respectively.
As 0.035 is lower than 0.05, $(K{{O}_{2}})$ is the limiting reagent and the product will be formed according to the number of moles of$(K{{O}_{2}})$. Water is an excess reagent because the given moles are greater than the moles of$(K{{O}_{2}})$.
So the number of moles produced of ${{O}_{2}}$ will be $=\dfrac{3}{4}\times 0.158=0.1185$ mol.

Hence the correct answer is option (A) i.e. moles of ${{O}_{2}}$ can be produced is 0.1185 mol.

Note:
Number of moles is calculated as the ratio of given mass of the substance to the molar mass of the substance. Don’t forget to balance the chemical reaction in such types of questions.