1. Ignoring any side reactions and assuming the reaction occurs completely, how

Question

1. 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.

2. At 1 atmosphere and 25 o C, what would the volume of O2 be from the previous question?    

3. Assuming that air is 21% O2 by volume, what volume of air would be needed to answer question 2?

4. Assuming all of the analogous reactions occur, why would lithium chlorate be a more practical choice for the “chlorate candle” than either sodium or potassium chlorate?

I got 17.2 kg KClO3 for the first one but don't know where to go from there.

Explanation / Answer

2KClO3(s) ----> 2Kcl(s) + 3O2(g)

mass of O2 required = 0.84*8 = 6.72 kg

   = 6720 grams

No of moles of O2 = (6720/32) = 210 mole.

so that,

No of moles of KClO3 required = 210*2/3 = 140 mole

molarmass of KClO3 = 122.55 g/mol

mass of KClO3 required = 140*122.55 = 17.157 Kg

2. v = volume of O2 = nRT/P

n= no of moles of O2 = 210 mole

R = gas constant = 0.0821 l.atm.k-1.mol-1

T = 25 c = 298 k

p = 1 atm

= 210*0.0821*298/1 = 5137.818 L

3. volume of air needed = 5137.818*100/21 = 24465.8 L

4. lithium chlorate is less ionic compared to others.so that it can be easily dissociates into O2 molecules.

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