Answer:
[tex]n_{CO_2}=1.93 gCO_2[/tex]
Explanation:
Hello,
In this case, considering the given chemical reaction, we can use the molar mass of octane (114.23 g/mol) and the 2:16 molar ratio with carbon dioxide to compute the emitted moles of CO2 to the atmosphere via the following stoichiometric procedure:
[tex]n_{CO2}=27.6gC_8H_{18}*\frac{1molC_8H_{18}}{114.23gC_8H_{18}} *\frac{16molCO_2}{2molC_8H_{18}} \\\\n_{CO_2}=1.93 gCO_2[/tex]
Which also corresponds to the following mass:
[tex]m_{CO_2}=1.93molCO_2*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=85.0gCO_2[/tex]
Best regards.
