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Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

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

Options:

Answer:

Solution:

Moles of $N_{2} = \frac{28}{28} = 1$ ,
Moles of $H_{2} = \frac{6}{2} = 3$
Moles of $N H_{3} = \frac{17}{17} = 1$

Moles of $N_{2} = \frac{1}{2} = 14$ g
Moles of $H_{2} = \frac{3}{2} \Rightarrow \frac{3 \times 2}{2} = 3 g$ .

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Mole ConceptSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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