Question

One of the most desirable of the old British sports cars was the beautiful Triumph Vitesse (1963-1971). Pictured below is the American version, called the Sports 6. A little expensive back in the day, only 679 were sold. Imagine you are on vacation, and for fun you and your companion select a cool, beautiful day in Gansu Province and ride the G30 highway from Lanzhou to the Mogao caves in Dunhuang, about distance of 686 miles. If you consume gasoline according to:
2 C8H18(l) + 25 O2(g)? 16 CO2(g) + 18 H2O(g)what volume of carbon dioxide gas would be produced by this motoring trip if your fuel consumption was 21.2 miles per gallon? Note that the density of gasoline is 0.805g/cm3, and one mole of any gas at 760 mmHg and 0oC is 22.4 L.

Answer 1

Answer:

$V_{CO_2}=1.55x10^{5}LCO_2=155m^3CO_2$

Explanation:

Hello,

In this case, given the reaction:

$2 C_8H_{18}(l) + 25 O_2(g)\rightarrow 16 CO_2(g) + 18 H_2O(g)$

The total consumed gallons are computed by considering 686 miles were driven and the consumption is 21.2 miles per gallon, thus:

$V_{C_8H_{18}}=686miles*\frac{1gal}{21.2miles} =32.4gal$

Hence, with the given density, one could compute the consumed grams and consequently moles of gasoline as well as moles that were consumed:

$n_{C_8H_{18}}=32.4gal*\frac{3785.41cm^3}{1gal} *\frac{0.805g}{1cm^3} *\frac{1mol}{114g}=864.95mol C_8H_{18}$

Next, since gasoline (molar mass = 114 is in a 2:16 molar relationship with the yielded carbon dioxide, we compute its produced moles as shown below:

$n_{CO_2}=864.95mol C_8H_{18}*\frac{16molCO_2}{2molC_8H_{18}} =6919.6molCO_2$

Finally, we could assume the given STP conditions to compute the volume of carbon dioxide, as no more information regarding the space wherein the carbon dioxide is available:

$V_{CO_2}=\frac{n_{CO_2}RT}{P} =\frac{6919.6mol*0.082\frac{atm*L}{mol*K}*(0+273)K}{1atm} \\\\V_{CO_2}=1.55x10^{5}LCO_2=155m^3CO_2$

Best regards.