According to the data booklet, propane releases 50.5 kj of energy per gram and 2220 kj of energy per mole.
Heat of combustion: 50.5 kj g-1
Molar heat of combustion: 2220 kj mol-1
To find out how much propane is consumed per minute we would divide 22.20kj (output of the stove per minute) by the heat of combustion and/or by molar heat of combustion. So:
22.20 kj/ 50.5 kj g-1 = 0.4396 g (0.440 g)
22.20 kj/ 2220 kj mol-1 = 0.01 mol
From this we can see that the answer is B.
For argument's sake, let's go through each answer.
For A:
Per minute, 0.01 mol of propane is consumed. The molar ratios tell us that per mole of propane, 5 mol of oxygen is consumed. Hence in this case, that's 0.05 mol of oxygen consumed.
To find the grams of oxygen we use the formula m(O2) = n x M
m(O2) = 0.05 mol x 16 g mol-1
= 0.8 g
Therefore A is wrong.
For C:
Per mole of propane, 3 mole of carbon dioxide is produced. Hence, 0.03 mol of carbon dioxide is produced per minute.
Hence we use the equation N = n x NA
N = 0.03 mol x 6.02 x 1023
= 1.81 x 1022
Therefore C is wrong
For D:
Per mole of propane, 4 mol of water is produced. Hence, 0.04 mol of water is produced per minute
We use the equation m = n x M
m(H2O) = 0.04 mol x 18 g mol-1
= 0.72 g
Hence D is wrong, due to wrong units