Ignoring any side reactions and assuming the reaction occurs completely, how lar
ID: 951871 • Letter: I
Question
Ignoring any side reactions and assuming the reaction occurs completely, how large (in kg) an oxygen candle (KCIO_3) 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 degreeC, what would the volume of O_2 be from the previous question? 3. Assuming that air is 21% O_2 by volume, what volume of air would be needed to answer question 2? 4. Assuming all of the analogous reactions occur, why would llithium chlorate be a more practical choice for the "chlorate candle" than either sodium or potassium chlorate?Explanation / Answer
The decomposition of KClO3 can be written as
2KClO3--à 2KCl+3O2
Molecular weights : KClO3=122.5 KCl= 74.5 O2= 32
Moles= Mass/Molecular weight
Oxygen needs to be supplied for 8 people/day= 0.84*8 kg/day=6.72 kg/day= 6.72*1000g/day=6720g/day
Converting this into moles give moles of oxygen =6720/32 gmoles/ day=210gmoles/day
From the stoichiometry
3moles of oxygen requires 2 moles of KClO3
210moles oxygen requires 210*2/3 =140moles of KClO3
Converting this into mass gives 140*122.5=17150gms/day
Mass o f KClO2= 17150/1000= 17.15kg/day
Since air contains 21% oxygen, moles of air to be supplied= 210/0.21=1000moles/day
From gas law equation, PV= nRT, volume of air can be calculated as
V= nRT/P , n= moles of air, R= Gas law constant= 0.08206L.atm/mole.K T= 25+273.15= 298.15K
P= 1atm
V= nRT/P =1000*0.08206L.atm/mole.K*(25+273.15)/1=24466.19 L
3. Since the molecular weight of LIClO3 is onle 90 compared to either sodium chlorate (whose molecular weight is 106) or KClO3( whose molecualr weight is 122.5), less amount of LiClO3 is requirs for producing same quantity of O2.