Consider the reaction sequence from \(\mathbf{P}\) to \(\mathbf{Q}\) shown below. The overall yield of the major product \(\mathbf{Q}\) from \(\mathbf{P}\) is \(75 \%\). What is the amount in grams of \(\mathbf{Q}\) obtained from 9.3 mL of \(\mathbf{P}\) ? (Use density of \(\mathbf{P}=1.00 \mathrm{~g} \mathrm{~mL}^{-1}\), Molar mass of \(\mathrm{C}=12.0, \mathrm{H}=1.0, \mathrm{O}=16.0\) and \(\mathrm{N}=14.0 \mathrm{~g} \mathrm{~mol}^{-1}\) )

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Compounds Containing Nitrogen

Consider the reaction sequence from

P

Q

shown below. The overall yield of the major product

Q

from

P

75%

. What is the amount in grams of

Q

obtained from 9.3 mL of

P

? (Use density of

P = 1.00 g mL^{- 1}

, Molar mass of

C = 12.0, H = 1.0, O = 16.0

and

N = 14.0 g mol^{- 1}

)

Answer: 18.6

Solution

Step 1: Calculate moles of P (aniline,

C_{6} H_{5} NH_{2}

)

Molar mass of aniline = 72 + 7 + 14 = 93 g/mol

Mass of P = 9.3 mL × 1.0 g/mL = 9.3 g

Moles of P = 9.3/93 = 0.1 mol

Step 2: The reaction sequence is 1:1:1 stoichiometry:

PhNH_{2} \to PhN_{2}^{+} \to azo dye Q

Theoretical moles of Q = 0.1 mol

With 75% yield: actual moles = 0.1 × 0.75 = 0.075 mol

Step 3: Calculate mass of Q (

C_{16} H_{12} ON_{2}

)

Molar mass of Q = 16(12) + 12(1) + 16 + 2(14) = 192 + 12 + 16 + 28 = 248 g/mol

Mass of Q = 0.075 × 248 = 18.6 g

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Compounds Containing Nitrogen

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