Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptCalculation of moleMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

An 81 gm mixture of MgCO₃(s) and NH₂COONH₄(s) in an equimolar ratio is heated in the open to a constant mass.

MgCO₃(s) $Δ$ MgO(s) + CO₂(g)

NH₂COONH₄(s) $Δ$ 2NH₃(g) + CO₂(g)

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

Answer:

Solution:

$MgCO_{3} (s) (a mole) - \to - - MgO (s) + CO_{2} (g) - (a mole)$

$NH_{2} COONH_{4} (s) (a mole) - \to - (2 a mole) 2 NH_{3} (g) + CO_{2} (g) - (a mole)$

$84 a + 78 a = 81$

$a = 0.5$

$Mass % MgCO_{3} = \frac{0.5 \times 84}{81} \times 100 = 51.9%$

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Calculation of mole

⏱ 2mℹ️ Source: QB

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