From the combustion of octane, the formaldehyde will be formed as this equation:
C8H18 + O2 → CH2O + H2O this is the original equation but it is not a balanced equation, so let's start to balance it:
the equation to be balanced so the number of atoms on the right side of the equation sholud be equal with the number of atoms on the lef side.
-we have 8 C atoms on left side and 1 atom on the right side so we will try putting 8 CH2O on the right side instead of CH2O
C8H18 + O2 → 8 CH2O + H2O
we have 2 O atoms on the left side and 9 atoms on the right side so we will try first to put 9 O2 instead of O2 on the left side and put 2H2O on the right side and put 16 CH2O instead of 8 CH2O to make the atoms of O are equal on both sides = 18 atoms
C8H18 + 9 O2 → 16 CH2O + 2H2O
put now we have 8 atom C on the left side and 16 atom on the right side so, we will put 2 C8H18 instead of C8H18 now we get this equation:
2C8H18 + 9O2 →16 CH2O + 2H2O
-now we have 36 of H atoms on both sides.
- and 16 of C atoms on both sides.
- and 18 of O atoms on both sides.
now all the number of atoms of O & C & H are equal on both sides
∴ 2C8H18 + 9O2 → 16 CH2O + 2 H2O
is the final balanced equation for the formation of formaldehayde
Answer:
The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = Heat gained = ?
c = Specific heat = 
ΔT = The change in temperature = 3.08°C
Now put all the given values in the above formula, we get:
Now we have to calculate molar enthalpy of combustion of this substance :
where,
= enthalpy change = ?
q = heat gained = 8.2544kJ
n = number of moles methane = 
Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Moles of lead(Pb) = 1.6x10^23/6.02x10^23 = 0.265 moles.
Weight of lead = moles x atomic weight of lead
= 0.265x207.2
= 54.908 grams.
Hope this helps!
