An emergency breathing apparatus often placed in mines works via the following chemical reaction: 4KO2(s) + 2CO2(g) to make 2K2CO3(s) + O2(g)

Question

If the oxygen supply becomes limited or if the air becomes poisoned, a worker can use the apparatus to breath while exiting the mine. Notice that the reaction produces O2, with 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 15 minutes to exit in an emergency? Assume that an adult consumes approximately 4.4g of oxygen in 15 minutes of normal breathing.

remember question?

you said to find the moles of O2 which I think is 1.1 mol O2
but I got confused on the second part.
"it takes four times that amount of moles of O2 of KO2. What is the mass of that?" do I find the moles of that and then that is it?

I just noticed you equation is not balanced. The O on the left don't match the right.

4KO2 + 2CO2 >> 2K2CO3 + 3O2

Does that balance?

Now, if you want say 4.4 grams of O2, convert that to moles, then you need 4/3 of that much (4/3) moles of potassium superoxide KO2.
amountKO2= gramsO2/molmassO2 * 4/3*molmassKO2.
or 4.4g/32g * 4/3 * (40 + 32) check that mole mass of KO2.
= 4.4*4*72/32 grams KO2
The entire point of this is that the balanced equation is the guide in the mole ratios of the reactants and products.