1. Write the combustion reaction

For the fuel $C_{15}H_{30}$:

$C_{15}H_{30} + O_2 \rightarrow CO_2 + H_2O$

Balancing carbon and hydrogen first:

$C_{15}H_{30} + O_2 \rightarrow 15CO_2 + 15H_2O$

Now balance oxygen on the right:

• In $15CO_2$, oxygen atoms $= 15 \times 2 = 30$
• In $15H_2O$, oxygen atoms $= 15 \times 1 = 15$

Total oxygen atoms required $= 45$, so:

$C_{15}H_{30} + \frac{45}{2}O_2 \rightarrow 15CO_2 + 15H_2O$

Thus, for $1$ mole of fuel:

• $\dfrac{45}{2}$ moles of $O_2$ are required
• $15$ moles of $CO_2$ are produced

2. Mass of 1 litre of fuel

Given density of fuel:

$0.756\ \text{g mL}^{-1}$

Since:

$1\ \text{L} = 1000\ \text{mL}$

Mass of $1$ litre fuel:

$m = 0.756 \times 1000 = 756\ \text{g}$

3. Moles of fuel in 1 litre

Molar mass of $C_{15}H_{30}$:

$15(12) + 30(1) = 180 + 30 = 210\ \text{g mol}^{-1}$

So moles of fuel:

$n = \frac{756}{210} = 3.6\ \text{mol}$

4. Oxygen required

From the balanced equation, $1$ mole fuel requires $\dfrac{45}{2} = 22.5$ moles $O_2$.

So for $3.6$ moles fuel:

$n(O_2) = 3.6 \times 22.5 = 81\ \text{mol}$

Mass of oxygen required:

$m(O_2) = 81 \times 32 = 2592\ \text{g}$

5. Carbon dioxide released

From the balanced equation, $1$ mole fuel gives $15$ moles $CO_2$.

So for $3.6$ moles fuel:

$n(CO_2) = 3.6 \times 15 = 54\ \text{mol}$

Mass of carbon dioxide released:

$m(CO_2) = 54 \times 44 = 2376\ \text{g}$

6. Match with options

Required values are:

• Oxygen required $= 2592\ \text{g}$
• $CO_2$ released $= 2376\ \text{g}$

This matches Option C.

7. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

So, they agree.