If a rocket runs on fuel C15H30 and liquid oxygen, the weight of O2 required and CO2 released for every litre of fuel…

🧠 Density $\to$ fuel mass $\to$ moles $\to$ products Reaction: $\ce{C15H30 + \tfrac{45}{2} O2 -> 15 CO2 + 15 H2O}$. So per mole of fuel: $22.5\,\text{mol}$ $\ce{O2}$ and $15\,\text{mol}$ $\ce{CO2}$.

🗺️ Four steps Mass of $1\,\text{L}$ fuel $= 1000 \times 0.756 = 756\,\text{g}$. Moles of $\ce{C15H30}$ ($M = 210$) $= 756 / 210 = 3.6\,\text{mol}$. Moles of $\ce{O2}$ needed $= 3.6 \times 22.5 = 81\,\text{mol}$. Mass $= 81 \times 32 = 2592\,\text{g}$. Moles of $\ce{CO2}$ produced $= 3.6 \times 15 = 54\,\text{mol}$. Mass $= 54 \times 44 = 2376\,\text{g}$.

So the pair is $(2592\,\text{g}\ \ce{O2},\ 2376\,\text{g}\ \ce{CO2})$.

⚡ Speed scan $\ce{O2}$ needed always exceeds $\ce{CO2}$ released for hydrocarbons (extra $\ce{O2}$ also makes water). So the first number must be the larger one. Only (c) fits.

$\boxed{\text{Answer: (c)}}$