Using your cars miles per gallon and the miles you drive in 1 week calculate the liters of carbon dioxide you produce when using your car. Compare your miles in February (pre-covid) versus June (post-covid). Assume that your engine combusts at 1.00 atm and 25.0 celsius. Assume that your gasoline is 100% octane (C8H18). You will also need to balance the combustion reaction that occurs in your car engine. You will need to use PV=nRT I think calculations are not correct specially for June. please help 2C8H18+ 25O2→16CO2+18H2O Feb, (pre covid), Miles covered per week=15×7=105 miles So fuel consumed per week=105 miles/30 miles ga =3.5 ga Density of octane=3.18 kg/ga Mass=volume × density Mass of Octane used per week = 3.5 ga x 3.18 kg/ga =11.13 kg Moles of octane in 11.13 kg 11.13 kg/114 kg/k mol =0.097 kmol 2 kmol octan≡16 kmol CO2 0.097 kmol octane =16/2 × 0.097 kmol CO2 0.097 kmol octane = 0.781 kmol CO2 n= 0.781 kmol=781 moles CO2 V=781 mol x 8.314j/k. mol x (25+273)K/1atm x 101 kpa V= 19.158L Final result in February car produced 19.158 L of CO2 per week in February (pre covid) June (post covid) Miles covered per week = 5 miles I used data I provided for February, 105 miles = 19.158 L CO2 5 miles distance = 19.158/105 x 5 L CO2 5 miles distance = 0.91228 L CO2 Final result for June car produced 0.9122 L of CO2 per week
Using your cars miles per gallon and the miles you drive in 1 week calculate the liters of carbon dioxide you produce when using your car. Compare your miles in February (pre-covid) versus June (post-covid). Assume that your engine combusts at 1.00 atm and 25.0 celsius. Assume that your gasoline is 100% octane (C8H18). You will also need to balance the combustion reaction that occurs in your car engine. You will need to use PV=nRT
I think calculations are not correct specially for June. please help
2C8H18+ 25O2→16CO2+18H2O
Feb, (pre covid),
Miles covered per week=15×7=105 miles
So fuel consumed per week=105 miles/30 miles ga =3.5 ga
Density of octane=3.18 kg/ga
Mass=volume × density
Mass of Octane used per week = 3.5 ga x 3.18 kg/ga =11.13 kg
Moles of octane in 11.13 kg
11.13 kg/114 kg/k mol =0.097 kmol
2 kmol octan≡16 kmol CO2
0.097 kmol octane =16/2 × 0.097 kmol CO2
0.097 kmol octane = 0.781 kmol CO2
n= 0.781 kmol=781 moles CO2
V=781 mol x 8.314j/k. mol x (25+273)K/1atm x 101 kpa
V= 19.158L
Final result in February car produced 19.158 L of CO2 per week in February (pre covid)
June (post covid)
Miles covered per week = 5 miles
I used data I provided for February, 105 miles = 19.158 L CO2
5 miles distance = 19.158/105 x 5 L CO2
5 miles distance = 0.91228 L CO2
Final result for June car produced 0.9122 L of CO2 per week
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