If a rocket runs on a fuel (C₁₅H₃₀) and liquid oxygen, the weight of oxygen required and CO₂ released for every litre of fuel respectively are :

(Given : density of the fuel is 0.756 g/mL)

<p>C₁₅H₃₀ + ${{45} \over 2}$O₂ $\to$ 15CO₂ + 15H₂O</p> <p>Given, volume of fuel = 1L = 1000 ml</p> <p>And density of fuel = 0.756 g/ml</p> <p>We know,</p> <p>$d = {w \over v}$</p> <p>$\Rightarrow 0.756 = {w \over {1000}}$</p> <p>$\Rightarrow$ w = 756 gm</p> <p>$\therefore$ weight of fuel = 756 gm</p> <p>Molar mass of C₁₅H₃₀ = 15 $\times$ 12 + 30 = 210</p> <p>$\therefore$ Moles of C₁₅H₃₀ = ${{756} \over {210}}$</p> <p>From equation you can see,</p> <p>1 mole of C₁₅H₃₀ react with ${{45} \over 2}$ mole of O₂</p> <p>$\therefore$ ${{756} \over {210}}$ moles of C₁₅H₃₀ react with ${{45} \over 2} \times {{756} \over {210}}$ moles of O₂</p> <p>$\therefore$ Moles of O₂ required = ${{45} \over 2} \times {{756} \over {210}}$</p> <p>$\therefore$ Mass of O₂ required = ${{45} \over 2} \times {{756} \over {210}}$ $\times$ 32 = 2592 g</p> <p>Also,</p> <p>From 1 mole of C₁₅H₃₀ 15 moles of CO₂ formed</p> <p>$\therefore$ From ${{756} \over {210}}$ moles of C₁₅H₃₀ 15 $\times$ ${{756} \over {210}}$ moles of CO₂ formed</p> <p>$\therefore$ Moles of CO₂ formed = 15 $\times$ ${{756} \over {210}}$</p> <p>$\therefore$ Mass of CO₂ formed = 15 $\times$ ${{756} \over {210}}$ $\times$ 44 = 2376 g</p>