Question 1 Report
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 3 of CO2 2 and 54g of water. The hydrocarbon should be
Answer Details
In the question above an Hydrocarbon combust to give CO2 and H20
Let Hydrocarbon beCxHy + x+Y/4O2= xCO2 + Y/2H2O
Mass of C0=44g and H2O=18gat STP vol= 22.4Therefore, 1mole of CO2 contains 44g and 22.4dm³ at STP
1mole = 22.4dm³xmole = 89.6dm³Cross multiplying x=89.6/22.4 =4mole of CO2 produce
1mole of H2O = 18gXmole = 56gCross multiplyingX = 56/18 = 3mole of H20Then....CxHy + X + y/4O2 = 4CO2+ 3H2O
BalancingC4H6 + 6O2 = 4CO2 + 3H2O
The IUPAC name for CH3 3 CH2 2 COOCH2 2 CH3 3 is
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