One would need #3498.9# grams of #O_2# for this particular combustion.
The combustion reaction can be written as follows
#C_9H_20 + 14O_2 -> 9CO_2 + 10H_2O#
we can see that #1# mole of nonane needs #14# moles of #O_2# for the combustion reaction; from #m_(C_9H_20)# = #1# kg, one can solve for the number of moles of nonane
#n_(C_9H_20) = (1000g)/(128 g/(mol)) = 7.81 # moles
(using #1 kg# = #1000 grams# and knowing the molar mass of nonane to be #128g/(mol)# );
Therefore, the number of #O_2# moles will be
#n_(O_2) = 14 * 7.81 = 109.34# moles
The mass of #O_2# yields
#m_(O_2) = 109.34 * 32 = 3498.9# grams
IF you need the quantity of #AIR#, just remember that air contains #20.95% O_2# , #78.09% N#, and a little under #1%# other gasses (#Ar, CO_2#,and others).
