Question

An emergency breathing apparatus often placed in mines or caves works via the following chemical reaction:
4KO2(s)+2CO2(g)→2K2CO3(s)+3O2(g) If the oxygen supply becomes limited or if the air becomes poisoned, a worker can use the apparatus to breathe while exiting the mine. Notice that the reaction produces O2, which can be breathed, and absorbs CO2, a product of respiration.

What minimum amount of KO2 is required for the apparatus to produce enough oxygen to allow the user 30 minutes to exit in an emergency? Assume that an adult consumes approximately 8.4 g of oxygen in 30 minutes of normal breathing.

Answer 1

The balanced chemical equation is -

4KO2(s)+2CO2(g)→2K2CO3(s)+3O2(g)

So, clearly 3 moles of produced by every 4 moles of KO2, so, 1 moles of O2 will need 4/3 moles of KO2 and thus x moles of O2 will need 4x/3 moles of KO2

Also. moles in 8.4g of O2 = weight/molar mass = 8.4g/ 32g/mole = 0.2625 moles

so, x = 0.2625 moles

So, required no. of moles of KO2 = 4x/3 moles

= 4*0.2625/3 moles = 0.35 moles

molar mass of KO2 = molar mass of K + molar mass of O2

= 39.09g/mole + 32g/mole = 71.09g/mole

So, weight of 0.35 moles of KO2 = molar mass*moles

= 71.09g/mole*0.35 moles = 24.8815 g

So, we will need 24.8815 g of KO2 every 30 mins to produce 8.4 g of O2 per 30 minutes