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

Answer:

- From octane: $m_{CO_2}=1.54x10^5gCO_2$

- From ethanol: $m_{CO_2}=9.57x10^4gCO_2$

Explanation:

Hello,

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

$C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O$

Thus, the produced mass of carbon dioxide is:

$m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2$

Now, for ethanol:

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

$m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2$

Best regards.