Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptLaws of Chemical CalculationsMedium2 minQB

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Passage / Comprehension

81gm mixture of $MgCO_{3} (s)$ and $NH_{2} COONH_{4} (s)$ in equimolar ratio is heated in open vessel to constant mass.

$MgCO_{3} (s) Δ MgO (s) + CO_{2} (g) NH_{2} COONH_{4} (s) Δ 2 NH_{3} (g) + CO_{2} (g)$

Mole % of $CO_{2}$ in the resulting gasesous mixture is

Answer:

Solution:

Let the moles of MgCO₃ and NH₂COONH₄ be $1 : 1$ .

MgCO₃ → MgO + CO₂, so 1 mole CO₂ is produced.

NH₂COONH₄ → 2NH₃ + CO₂, so 1 mole CO₂ is produced.

Total gaseous moles = CO₂ (2) + NH₃ (2) = 4 moles.

Mole % of CO₂ = $(2/4) \times 100 = 50%$ .

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

⏱ 2mℹ️ Source: QB

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