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.

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

Mass % of $MgCO_{3}$ in original sample -

Answer:

Solution:

MgCO₃ + NH₂COONH₄ (equimolar ratio) taken.

Let moles of each = 1 mole. Mass of MgCO₃ = 84 g, NH₂COONH₄ = 78 g.

On heating: MgCO₃ → MgO + CO₂ and NH₂COONH₄ → 2NH₃ + CO₂ + H₂O.

Remaining mass = MgO (40 g) + gases removed from second compound (no residue) = 40 g.

Original mass = 84 + 78 = 162 g, so % MgCO₃ = $(84/162) \times 100 = 51.9%$ .

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

⏱ 2mℹ️ Source: QB

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