C₁₅H₃₀ + $${{45} \over 2}$$O₂ $$\to$$ 15CO₂ + 15H₂O

Given, volume of fuel = 1L = 1000 ml

And density of fuel = 0.756 g/ml

We know,

$$d = {w \over v}$$

$$ \Rightarrow 0.756 = {w \over {1000}}$$

$$\Rightarrow$$ w = 756 gm

$$\therefore$$ weight of fuel = 756 gm

Molar mass of C₁₅H₃₀ = 15 $$\times$$ 12 + 30 = 210

$$\therefore$$ Moles of C₁₅H₃₀ = $${{756} \over {210}}$$

From equation you can see,

1 mole of C₁₅H₃₀ react with $${{45} \over 2}$$ mole of O₂

$$\therefore$$ $${{756} \over {210}}$$ moles of C₁₅H₃₀ react with $${{45} \over 2} \times {{756} \over {210}}$$ moles of O₂

$$\therefore$$ Moles of O₂ required = $${{45} \over 2} \times {{756} \over {210}}$$

$$\therefore$$ Mass of O₂ required = $${{45} \over 2} \times {{756} \over {210}}$$ $$\times$$ 32 = 2592 g

Also,

From 1 mole of C₁₅H₃₀ 15 moles of CO₂ formed

$$\therefore$$ From $${{756} \over {210}}$$ moles of C₁₅H₃₀ 15 $$\times$$ $${{756} \over {210}}$$ moles of CO₂ formed

$$\therefore$$ Moles of CO₂ formed = 15 $$\times$$ $${{756} \over {210}}$$

$$\therefore$$ Mass of CO₂ formed = 15 $$\times$$ $${{756} \over {210}}$$ $$\times$$ 44 = 2376 g