Chemistry - Hydrocarbon Question with Solution | TestHub

ChemistryHydrocarbonCHEMICAL REACTIONS OF ALKYNESMedium2 minPYQ_2021

ChemistryMediumnumerical

Passage / Comprehension

For the following reaction scheme, percentage yields are given along the arrow:

$x g$ and $y g$ are mass of $R$ and $U$ , respectively. (Use: Molar mass (in $g mol^{- 1}$ ) of $H, C$ and $O$ as 1,12 and 16 , respectively)

The value of $x$ is ___.

Answer:

1.62

Solution:

$M g_{2} C_{3} H_{2} O$ C₃H₄(P)Propyne (4g) + Mg(OH)₂

Now, one mole of propyne give, one mole of $2$ -butyne. $3$ moles of $2$ -butyne give one mole of hexamethyl benzene.

Hence, $40 g$ of propyne give $54 g$ to $2$ -butyne $(100 %)$ but given that $75 %$ yeild and $4 g$ to propyne.

The mass of $2$ -butyne formed $= 5 .4 \times \frac{75}{100} = 4 .05 g$ $162 g$ to $2$ -butyne give $162 g$ of hexamethyl benzene $(100 % yield)$

But given that $40 %$ yield, and $4.05 g$ of $2$ -butyne

The mass of hexamethyl benzene formed $= 4 .05 \times \frac{40}{100} = 1 .62 g (x)$

Stream:JEE_ADVSubject:ChemistryTopic:HydrocarbonSubtopic:CHEMICAL REACTIONS OF ALKYNES

⏱ 2mℹ️ Source: PYQ_2021

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