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ICSE Class 10 Chemistry Question 60 of 72

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 60

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Question

Question 50(b-i)

67.2 litres of hydrogen combines with 44.8 litres of nitrogen to form ammonia :

N2(g) + 3H2(g) ⟶ 2NH3(g).

Calculate the vol. of ammonia produced. What is the substance, if any, that remains in the resultant mixture ?

Answer

[By Lussac's law]

N2+3H2⟶2NH31 vol.:3 vol.⟶2 vol.\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2 \text{NH}_3 \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}

To calculate the volume of ammonia gas formed.

H2:NH33:267.2:x\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 67.2 & : & x \end{matrix}

∴23×67.2=44.8 lit\therefore \dfrac{2}{3} \times 67.2 = 44.8 \text{ lit}

Hence, volume of NH3 formed is 44.8 lit.

We know,

H2:N23:167.2:x\begin{matrix}\text{H}_2 & : & \text{N}_2 \\ 3 & : & 1 \\ 67.2 & : & x \end{matrix}

∴13×67.2=22.4 lit\therefore \dfrac{1}{3} \times 67.2 = 22.4 \text{ lit}

∴ nitrogen left = 44.8 - 22.4 = 22.4 lit.

Hence, 22.4 lit of nitrogen remains in the resultant mixture.