Ignoring any side reactions and assuming the reaction occurs completely, how lar
ID: 820300 • Letter: I
Question
Ignoring any side reactions and assuming the reaction occurs completely, how large (in kg) an oxygen candle (KClO3) would be needed to supply 8 people with enough oxygen for 24 hours on a small submarine? Although this depends on the size of the person and their respiration rate (activity), according to NASA, an average person needs about 0.84 kg of O2 per day. = 17.2 kg
At 1 atmosphere and 25 oC, what would the volume of O2 be from the previous question?
Assuming that air is 21% O2 by volume, what volume of air would be needed to answer question 2?
I have number one but have no idea how to do the second two questions
Explanation / Answer
.84kg/person x 8 people = 6.72kg
6.72kg x 1000g/kg = 6720g
6720g x (1 mole / 32 g O2 [molar mass of oxygen]) x (2 moles KClO3 / 3 moles O2 [ratio of moles KClO3 to moles of O2]) x (122.55g KClO3 / 1 Mole) = 17157g KClO3
17157g KClO3 x (1kg / 1000g) = 17.2kg