Answer:
![n_(CO_2)=1.93 gCO_2](https://img.qammunity.org/2021/formulas/chemistry/college/3qfityh06kv8ojaljythdhkw2gyqgjpnwp.png)
Step-by-step 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:
![n_(CO2)=27.6gC_8H_(18)*(1molC_8H_(18))/(114.23gC_8H_(18)) *(16molCO_2)/(2molC_8H_(18)) \\\\n_(CO_2)=1.93 gCO_2](https://img.qammunity.org/2021/formulas/chemistry/college/5doonrwhqkrmdpwm24ckmy2mtwnfspb0nm.png)
Which also corresponds to the following mass:
![m_(CO_2)=1.93molCO_2*(44gCO_2)/(1molCO_2) \\\\m_(CO_2)=85.0gCO_2](https://img.qammunity.org/2021/formulas/chemistry/college/u3gwzurjs5jnaagw011vv9zg1d4asl0dah.png)
Best regards.