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Redox Reactions

Potassium sup...

Question

Potassium superoxide ( ${K O}_{2}$ ) is used in rebreathing gas masks to generate oxygen:
${K O}_{2} (s) + H_{2} O (l) ⟶ K O H (s) + O_{2} (g)$
If a reaction vessel contains 0.158 mol ${K O}_{2}$ and 0.10 mol $H_{2} O$ , how many moles of $O_{2}$ can be produced?

0.1185

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0.1125

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0.1265

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0.2008

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Solution

The correct option is C 0.1185
The balanced reaction is given below:
$4 K O_{2} + 2 H_{2} O \to 4 K O H + 3 O_{2}$
$0.158$ ; $0.10$ ( No. of moles given)
$\frac{0.158}{4}$ ; $\frac{0.10}{2}$ ( Ratio of moles given and stoichiometry coefficient )
= $0.0395$ ; = $0.05$
As $0.035$ is lower than $0.05$ , $K O_{2}$ is limiting reagent and product will be formed according to number of moles of $K O_{2}$ (given).
So, moles produced of $O_{2} = \frac{3}{4} \times 0.158 = 0.1185$ mol.

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