Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptLaws of Chemical CalculationsMedium2 minQB

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28 $g N_{2}$ and $6 g H_{2}$ were mixed. At equilibrium $17 g NH_{3}$ was formed. The weight of $N_{2}$ and $H_{2}$ at equilibrium are respectively.

Answer:

Solution:

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$

moles of $t = 0$ : 1 3 0

Moles at equilibrium: $(1 - x)$ $(3 - 3 x)$ $2 x$

Given that $2 x = 1$ ⇒ $x = \frac{1}{2}$ .

Moles of $N_{2}$ at equilibrium = $1 - \frac{1}{2} = \frac{1}{2}$ mole

Mass of $N_{2}$ at equilibrium = $\frac{1}{2} \times 28 = 14 g m$

Moles of $H_{2}$ at equilibrium:

$3 - 3 (\frac{1}{2}) = 3 - \frac{3}{2} = \frac{3}{2}$ moles

Mass of $H_{2}$ at equilibrium:

$\frac{3}{2} \times 2 = 3 g m$

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Laws of Chemical Calculations

⏱ 2mℹ️ Source: QB

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