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Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptMiscellaneous/MixedMedium2 minPYQ_2020

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Aluminum carbide $(A l_{4} C_{3})$ liberates methane on treatment with water. The grams of aluminum carbide required to produce $11 .2 L$ of methane under STP conditions is [Given $Al = 27$ ]

Options:

Answer:

Solution:

$A l_{4} C_{3} + 12 H_{2} O \to 3 C H_{4} + 4 A l (O H_{3})$
So $67 .2 L$ $C H_{4}$ evolved by $144 g$ $A l_{4} C_{3}$
$11 .2 L$ $C H_{4}$ will be evolved $= \frac{144}{67.2} \times 11.2 = 24 g$

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

⏱ 2mℹ️ Source: PYQ_2020

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