potassium superoxide (ko₂) is used in rebreathing masks to produce oxygen and remove carbon dioxide.\n4…

potassium superoxide (ko₂) is used in rebreathing masks to produce oxygen and remove carbon dioxide.\n4 ko₂(s) + 2 co₂(g) → 2 k₂co₃(s) + 3 o₂(g)\nif 35.5 grams of potassium superoxide react completely with carbon dioxide at stp, what volume of oxygen gas will be produced at stp?

potassium superoxide (ko₂) is used in rebreathing masks to produce oxygen and remove carbon dioxide.\n4 ko₂(s) + 2 co₂(g) → 2 k₂co₃(s) + 3 o₂(g)\nif 35.5 grams of potassium superoxide react completely with carbon dioxide at stp, what volume of oxygen gas will be produced at stp?

Answer

Explanation:

Step1: Calculate moles of $KO_2$

The molar mass of $KO_2$ is $39 + 2\times16=71$ g/mol. Moles of $KO_2$ = $\frac{35.5}{71}=0.5$ mol.

Step2: Determine mole - ratio

From the balanced equation $4KO_2(s)+2CO_2(g)\rightarrow2K_2CO_3(s) + 3O_2(g)$, the mole - ratio of $KO_2$ to $O_2$ is $4:3$.

Step3: Calculate moles of $O_2$

Let the moles of $O_2$ be $x$. Using the mole - ratio $\frac{4}{3}=\frac{0.5}{x}$, we get $x = 0.375$ mol.

Step4: Calculate volume of $O_2$ at STP

At STP, 1 mole of any gas occupies 22.4 L. So the volume of $O_2$ is $V = 0.375\times22.4 = 8.4$ L.

Answer:

8.4 L