I really need help! Please only respond if you know what your doing for each par

Question

I really need help! Please only respond if you know what your doing for each part.

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 oC, 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?

Explanation / Answer

1. For this chlorate candle, we have the following reaction:

2KClO3 + heat -> 2KCl + 3O2

As we have 8 people and each one requires 0.84 kg per 24 hours, we have 0.84 * 24 = 20.16 kg

We convert those kg to kmol:

20.16 kg * (1kmol / 32 kg) = 0.63 kmol of O2

Now we convert them to kmol of KClO3 with stoichiometric relation:

0.63 kmol of O2 * (2 kmol of KClO3 / 3 kmol of O2) = 0.42 kmol of KClO3

We finally convert these kmol to kg:

0.42 kmol of KClO3 * (122.55 kg / kmol) = 51.471 kg of KClO3

2. Assuming ideal gas:

PV = nRT

1 atm * V = 630 mol * 0.082 L atm / K mol * 298 K

V = 15394.68 L

3. Volume of air needed:

15394.68 L / 0.21 = 73308 L

4. Because lithium candles will produce more oxygen per kg of candle, as lithium has a lower molar mass than sodium or potassium.

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