JEE Main 2022 — Some Basic Concepts Of Chemistry Question with Solution

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

Given, volume of fuel = 1L = 1000 ml

And density of fuel = 0.756 g/ml

We know,

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

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

$$\Rightarrow$$ w = 756 gm

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

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

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

From equation you can see,

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

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

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

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

Also,

From 1 mole of C₁₅H₃₀ 15 moles of CO₂ formed

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

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

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