When a hydrocarbon A undergoes combustion in the presence of air, it requires $$9.5$$ equivalents of oxygen and produces $$3$$ equivalents of water. What is the molecular formula of A?

We need to find the molecular formula of hydrocarbon A that requires 9.5 equivalents of O$$_2$$ for combustion and produces 3 equivalents of water.

Write the general combustion equation.

For a hydrocarbon C$$_x$$H$$_y$$:

$$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \to x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$$

Set up equations from the given data.

Moles of O$$_2$$ required: $$x + \frac{y}{4} = 9.5$$

Moles of H$$_2$$O produced: $$\frac{y}{2} = 3 \implies y = 6$$

Solve for $$x$$.

$$x + \frac{6}{4} = 9.5$$

$$x + 1.5 = 9.5$$

$$x = 8$$

Identify the molecular formula.

The molecular formula is C$$_8$$H$$_6$$.

The correct answer is Option A: C$$_8$$H$$_6$$.