Solve the following:

(i) What volume of oxygen is required to burn completely 90 dm³ of butane under similar conditions of temperature and pressure?

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

(ii) The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP?

(iii) A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?

(i) [By Lussac's law]

$\begin{matrix} 2\text{C}$4\text{H}{10} & + & 13\text{O}2 & \longrightarrow & 8\text{CO}2 & + & 10\text{H}_2\text{O} \ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} & : & 10\text{ vol.} \end{matrix}

To calculate the volume of oxygen

$\begin{matrix}\text{C}$4\text{H}{10} & : & \text{O}_2 \ 2 & : & 13 \ 90 & : & x \end{matrix}

∴

$\dfrac{13}{2} \times 90 = 585 \text{ dm}^3$

Hence, volume of oxygen required is 585 dm³

(ii) Given, V.D. = 8

Gram molecular mass = V.D. x 2 = 8 x 2 = 16 g.

16 g occupies 22.4 lit.

∴ 24 g. will occupy = $\dfrac{22.4}{16}$ x 24 = 33.6 lit.

Hence, volume occupied by gas = 33.6 lit.

(iii) According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Hence, number of molecules of N₂ = Number of molecules of H₂ = X