Question

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867 g/mL) is consumed when a sample of gasoline burns in air. How many grams of oxygen are needed for complete combustion of the toluene? (a) How many grams of oxygen are needed for complete combustion of the toluene? g (b) How many total moles of gaseous products form? mol (c) How many molecules of water vapor form?

Answer 1

Answer:

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene \times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} \\\\=4.49 \text { mol gaseous products (Answer) }$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene \times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} \\\\ =9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) }$