Answer:
- From octane: [tex]m_{CO_2}=1.54x10^5gCO_2[/tex]
- From ethanol: [tex]m_{CO_2}=9.57x10^4gCO_2[/tex]
Explanation:
Hello,
At first, for the combustion of octane, the following chemical reaction is carried out:
[tex]C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O[/tex]
Thus, the produced mass of carbon dioxide is:
[tex]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[/tex]
Now, for ethanol:
[tex]C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O[/tex]
[tex]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[/tex]
Best regards.
