Question #234055

octane has chemical formula of C8H18 and is the main component in gasoline .Assume a 1.5 litre 4 stroke 4 cylinder automobile engine running on octane .if the engine is running at 3000rpm and utilize all incoming air to combust octane while the car is moving at 120Km/h what is the fuel consumption of the car in km/l .1 l of air contain 0.25g of o2. .Density of octane is 700g/l. Take relative atomic masses as c=12, o= 16 and H=1.


1
Expert's answer
2021-09-07T02:08:54-0400

2C8H18+25O2>16CO2+18H2O2 C_8H_{18} + 25 O_2 --> 16 CO_2 + 18 H_2O


Molar mass of C8H18=8(12)+18(1)=114C_8H_{18}= 8(12)+18(1)=114



Molar mass of O2=2(16)=32O_2=2(16)=32



Moles of O2=0.2532=0.0078molesO_2=\frac{0.25}{32}=0.0078moles



Moles of octane =225×0.0078=0.00063moles=\frac{2}{25}×0.0078=0.00063moles


Mass of octane =0.00063×114=0.0713g=0.00063×114=0.0713g






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