Question

Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g of ethanol (C2H6O). How much carbon dioxide (CO2), in grams, is produced in one tank of gas from the combustion of each amount?

Answer 1

- From octane:
`m_(CO_2)=1.54x10^5gCO_2`

- From ethanol:
`m_(CO_2)=9.57x10^4gCO_2`

Step-by-step explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

`C_8H_(18)+(25)/(2) O_2\rightarrow 8CO_2+9H_2O`

Thus, the produced mass of carbon dioxide is:

`m_(CO_2)=5.00x10^4gC_8H_(18)*(1molC_8H_(18))/(114gC_8H_(18))*(8molCO_2)/(1molC_8H_(18))*(44gCO_2)/(1molCO_2) \\\\m_(CO_2)=1.54x10^5gCO_2`

Now, for ethanol:

`C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O`

`m_(CO_2)=5.00x10^4gC_2H_6O*(1molC_2H_6O)/(46gC_2H_6O)*(2molCO_2)/(1molC_2H_6O)*(44gCO_2)/(1molCO_2) \\\\m_(CO_2)=9.57x10^4gCO_2`

Best regards.