# Question 53a00

Dec 3, 2014

One would need $3498.9$ grams of ${O}_{2}$ for this particular combustion.

The combustion reaction can be written as follows

${C}_{9} {H}_{20} + 14 {O}_{2} \to 9 C {O}_{2} + 10 {H}_{2} O$

we can see that $1$ mole of nonane needs $14$ moles of ${O}_{2}$ for the combustion reaction; from ${m}_{{C}_{9} {H}_{20}}$ = $1$ kg, one can solve for the number of moles of nonane

${n}_{{C}_{9} {H}_{20}} = \frac{1000 g}{128 \frac{g}{m o l}} = 7.81$ moles

(using $1 k g$ = $1000 g r a m s$ and knowing the molar mass of nonane to be $128 \frac{g}{m o l}$ );

Therefore, the number of ${O}_{2}$ moles will be

${n}_{{O}_{2}} = 14 \cdot 7.81 = 109.34$ moles

The mass of ${O}_{2}$ yields

${m}_{{O}_{2}} = 109.34 \cdot 32 = 3498.9$ grams

IF you need the quantity of $A I R$, just remember that air contains 20.95% O_2 , 78.09% N, and a little under 1%# other gasses ($A r , C {O}_{2}$,and others).