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JEE Advanced 2021 Paper 1

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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)

Section: Chemistry

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Question : 24 of 57

Marks: +1, -0

The value of x is_____________.

[JEE Adv 2021 P1]

Your Answer:

Solution:

\mathrm{Mg}_2\mathrm{C}_3 + 4\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Mg}(\mathrm{OH})_2 + \mathrm{CH}_3\mathrm{C} \equiv \mathrm{CH}

3\mathrm{CH}_3-\mathrm{C}\equiv \mathrm{C}-\mathrm{CH}_3 \xrightarrow[\underset{(40\%)}{\text{iron tube 873 K}}]{\text{red hot}}

Given, amount of

P = 4 \, \text{g} = 0.1 \, \text{mol}

From

0.1 \, \text{mol}

of

P, \, 0.01 \, \text{mol}

of

R

will be produced.

So, the value of

x = 162 \times 0.01 = 1.62 \, \text{g}

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