Question

In self-contained breathing devices used by first responders, potassium superoxide, KO2, reacts with exhaled carbon dioxide to produce potassium carbonate and oxygen:

Unbalanced: KO2(s) + CO2(g) --> K2CO3(s) + O2(g)

How much O2 could be produced from 85 g KO2?

Answer 1

Approximately 19.152 grams of O2 could be produced from 85 g KO2.

Step-by-step explanation:

To calculate the amount of O2 produced from 85 g of KO2, we first need to determine the balanced equation for the reaction. From the given information, 2 mol of KO2 react with 1 mol of O2. The molar mass of KO2 is 71.10 g/mol, so for 85 g of KO2, we have:

(85 g KO2) / (71.10 g/mol KO2) = 1.197 mol KO2

Using the stoichiometry of the reaction, we can calculate the moles of O2 produced:

1.197 mol KO2 * (1 mol O2 / 2 mol KO2) = 0.5985 mol O2

Finally, we convert the moles of O2 to grams:

0.5985 mol O2 * (32 g/mol O2) = 19.152 g O2

Therefore, approximately 19.152 grams of O2 could be produced from 85 g KO2.

Answer 2

7.37. Oxygen for First Responders In self-contained breathing devices used by first responders, potassium superoxide, KO9, reacts with exhaled carbon dioxide to produce potassium carbonate and oxygen: 4 KO2s) 2 CO2g)2 K9CO3(s)+3 O2(g) How much O9 could be produced from 85g KOg?