QJEEM22Ch3e39: JEE Main 2022 Chemistry - Class 11 JEEM22Ch3e39 Solution: If a rocket runs on a fuel (C 15 H 30 ) and liquid oxygen, the…

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