Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptCalculation of moleMedium2 minQB

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The % loss in mass upon strong heating of $NaHCO_{3}$ is nearly :

Answer:

Solution:

$2 NaHCO_{3} (s) ⟶ N a_{2} CO_{3} (s) + H_{2} O (g) + CO_{2} (g)$

↓↓

18g + 44 g = 62

$2 (23 + 1 + 60) = 168 % l oss = \frac{62}{168} \times 100 = 36.9 ≅ 37%$

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

⏱ 2mℹ️ Source: QB

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