Question

When organic fuels like propane (C3H8) burn completely with oxygen from the air, all the carbon atoms in the propane molecules make carbon dioxide (CO2) molecules and all the hydrogen atoms make water (H2O) molecules.
C3H8 + 5 O2  3 CO2 + 4 H2O.
This is a balanced reaction because there are 3 carbon atoms on each side of the arrow, 8 hydrogen atoms on each side of the arrow, and 10 oxygen atoms on each side (5 O2 means five molecules with 2 Oxygen atoms each, for a total of ten on the left side. 3 CO2 means three molecules with two atoms of oxygen each, or six atoms, and 4 H2O means four molecules with one oxygen each, four more, for a total of ten on the right side.)

Balance the reaction for the complete combustion (burning) of pentane, C5H12. First, decide how many carbon dioxide molecules and how many water molecules will be made. Then count the total oxygen atoms on the right side and figure out how many O2 molecules will be needed to make that many oxygen atoms.

C5H12 + ___ O2  ___ CO2 + ___ H2O

The ethanol (C2H6O) which is added to gasoline burns the same way, except that one of the oxygen atoms which goes to the products is already in the ethanol molecule.
Balance the reaction for burning ethanol in the same way.

C2H6O   + ___ O2  ___ CO2 + ___ H2O

Now try balancing the reaction for the burning of the sugar glucose during respiration:

C6H12O6   + ___ O2  ___ CO2 + ___ H2O

Answer 1

ANSWER-

As we know ,

When organic fuels burn completely with oxygen from the air, all the carbon atoms in the organic molecules make carbon dioxide (CO₂) molecules and all the hydrogen atoms make water (H₂O) molecules.

example - C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

the above example is for alkane molecules having general formula C_nH_2n+2 i.e. C₃H_2x3+2

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since given reaction is also burning of alkane (C₅H_{5 x2 +2}) molecule so would follow the same trend as in the given example.

so there would be 5 molecules of CO₂ as there are 5 carbons and would be 6 molecules of H₂O since there are 12 hydrogens in the given alkane. There would be 8 molecules of O₂ .

C₅H₁₂ + 8 O₂ ----> 5 CO₂ + 6 H₂O

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Burning of Ethanol molecule-

since there is 2 carbons there would be 2 molecules of CO₂ and since there is 6 hydrogens so there would be 3 molecules of the water  in the given molecule .

C₂H₆O   + 3 O₂ ----->  2 CO₂ + 3 H₂O

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Consider the following reaction-

C₆H₁₂O₆   +6 O₂ -----> 6 CO₂ + 6 H₂O

since there is 6 carbons there would be 6  molecules CO₂ , 12 hydrogens so ,6 molecules of  H₂O, accordingly 6 molecules of  O₂

Generalised expression for carbohydrate burning is

C_nH_2nO_n +n O2 -----> n CO2 + n H₂O