Answer:
[tex]m_{CO_2}=2.8gCO_2[/tex]
Explanation:
Hello,
In this case, the combustion of octane is chemically expressed by:
[tex]C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O[/tex]
In such a way, due to the 25/2:8 molar ratio between oxygen and carbon dioxide, we can compute the yielded grams of carbon dioxide (molar mass 44 g/mol) as shown below:
[tex]m_{CO_2}=3.2gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{\frac{25}{2}molO_2 } *\frac{44gCO_2}{1molCO_2}\\ \\m_{CO_2}=2.8gCO_2[/tex]
Best regards.