Answer:
12.3g of [tex]CO_{2}[/tex]
Explanation:
1. First write the balanced chemical equation for the combustion of the octane:
[tex]_2C_{8}H_{18}+_{25}O_{2}=_{16}CO_{2}+_{18}H_{2}O[/tex]
2. Determine the limiting reagent between the octane and the oxygen:
To find the limiting reagent take the mass given of each compound and divide it by the molar mass of each one, then divide the result between the stoichiometric coefficient in the balanced chemical reaction.
The compound with the smallest number will be the limiting reagent.
- For the oxygen:
[tex]15.0g*\frac{1mol}{32g}= 0.47molesO_{2}[/tex]
[tex]\frac{0.47}{25}=0.019[/tex]
- For the octane:
[tex]15.0g*\frac{1mol}{114.23g}=0.11molesC_{8}H_{18}[/tex]
[tex]\frac{0.13}{2}=0.052[/tex]
As the smallest number is for the oxygen, it is the limiting reagent.
3. Calculate the mass of carbon dioxide produced.
[tex]15.0gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{16molesCO_{2}}{25molesO_{2}}*\frac{44gCO_{2}}{1molCO_{2}}=13.2gCO_{2}[/tex]
As the combustion engine gives a yield of 93%, the carbon dioxide produced will be:
[tex]13.2gCO_{2}*0.93=12.3gCO_{2}[/tex]
