Step 1: Write the combustion reaction.
The combustion of a hydrocarbon \( C_aH_b \) is given by the reaction:
\[
C_aH_b + O_2 \rightarrow CO_2 + H_2O.
\]
Step 2: Write the stoichiometric coefficients.
The number of moles of \( CO_2 \) produced is equal to the number of moles of carbon in the hydrocarbon, which is \( a \). The number of moles of \( H_2O \) produced is half the number of moles of hydrogen atoms in the hydrocarbon, which is \( \frac{b}{2} \).
Step 3: Apply the oxygen balance.
The oxygen atoms required for the reaction are from \( O_2 \), and they are needed for both \( CO_2 \) and \( H_2O \):
- \( CO_2 \) requires \( a \) moles of \( O_2 \) (1 mole of \( O_2 \) provides 2 oxygen atoms, and each \( CO_2 \) molecule needs 2 oxygen atoms).
- \( H_2O \) requires \( \frac{b}{2} \) moles of \( O_2 \).
Thus, the total oxygen requirement is:
\[
\text{Total } O_2 = \frac{a}{2} + \frac{b}{4}.
\]
But we are given that the total oxygen requirement is 7.5 moles of \( O_2 \), so:
\[
\frac{a}{2} + \frac{b}{4} = 7.5.
\]
Step 4: Use the moles of \( CO_2 \) produced.
The number of moles of \( CO_2 \) produced is 6, so \( a = 6 \).
Step 5: Solve for \( b \).
Substitute \( a = 6 \) into the oxygen equation:
\[
\frac{6}{2} + \frac{b}{4} = 7.5 \quad \Rightarrow \quad 3 + \frac{b}{4} = 7.5 \quad \Rightarrow \quad \frac{b}{4} = 4.5 \quad \Rightarrow \quad b = 18.
\]
So, the correct values of \( a \) and \( b \) are 6 and 6, respectively.