JEE Challenger
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ChemHardAdv 2021 • Paper 2

Mass of Nitrobenzene Converted to Organic Salt

Comprehension Passage

The quantitative reaction of x gx\text{ g} of Sn\text{Sn} with HCl\text{HCl} yields a salt. This entire amount of salt is subsequently reacted with y gy\text{ g} of nitrobenzene in the presence of required quantity of HCl\text{HCl} to quantitatively produce 1.29 g1.29\text{ g} of an organic salt. (Take molar masses of H, C, N, O, Cl\text{H, C, N, O, Cl}, and Sn\text{Sn} as 1,12,14,16,351, 12, 14, 16, 35, and 119 g mol−1119\text{ g mol}^{-1}, respectively).

Determine the value of yy.

Official Numerical Answer1.23

Topics & Concepts

AminesMole ConceptStoichiometryAminesOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the value of yy (the mass of nitrobenzene converted to the organic salt), we follow a step-by-step stoichiometric analysis based on the given quantitative chemical reactions.

Step 1: Identify the Organic Salt and Calculate its Molar Mass

The reduction of nitrobenzene (C6H5NO2\text{C}_6\text{H}_5\text{NO}_2) using tin (Sn\text{Sn}) and hydrochloric acid (HCl\text{HCl}) yields aniline (C6H5NH2\text{C}_6\text{H}_5\text{NH}_2). In the presence of excess HCl\text{HCl}, aniline forms an organic salt, anilinium chloride (C6H5NH3+Cl−\text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^-).

Using the atomic masses provided in the question:

  • C=12 g mol−1\text{C} = 12\text{ g mol}^{-1}
  • H=1 g mol−1\text{H} = 1\text{ g mol}^{-1}
  • N=14 g mol−1\text{N} = 14\text{ g mol}^{-1}
  • Cl=35 g mol−1\text{Cl} = 35\text{ g mol}^{-1}

The chemical formula of anilinium chloride is C6H8NCl\text{C}_6\text{H}_8\text{NCl}.

Molar mass of C6H8NCl=(6×12)+(8×1)+14+35=72+8+14+35=129 g mol−1\text{Molar mass of }\text{C}_6\text{H}_8\text{NCl} = (6 \times 12) + (8 \times 1) + 14 + 35 = 72 + 8 + 14 + 35 = 129\text{ g mol}^{-1}

Step 2: Calculate the Moles of Organic Salt Produced

Given the mass of the organic salt produced is 1.29 g1.29\text{ g}:

Moles of anilinium chloride=MassMolar mass=1.29 g129 g mol−1=0.01 mol\text{Moles of anilinium chloride} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.29\text{ g}}{129\text{ g mol}^{-1}} = 0.01\text{ mol}

Step 3: Calculate the Molar Mass of Nitrobenzene

The chemical formula of nitrobenzene is C6H5NO2\text{C}_6\text{H}_5\text{NO}_2.

Using the given atomic masses:

  • O=16 g mol−1\text{O} = 16\text{ g mol}^{-1}

Molar mass of C6H5NO2=(6×12)+(5×1)+14+(2×16)=72+5+14+32=123 g mol−1\text{Molar mass of }\text{C}_6\text{H}_5\text{NO}_2 = (6 \times 12) + (5 \times 1) + 14 + (2 \times 16) = 72 + 5 + 14 + 32 = 123\text{ g mol}^{-1}

Step 4: Determine the Mass of Nitrobenzene (yy)

Since 1 mole1\text{ mole} of nitrobenzene (C6H5NO2\text{C}_6\text{H}_5\text{NO}_2) quantitatively produces 1 mole1\text{ mole} of anilinium chloride (C6H8NCl\text{C}_6\text{H}_8\text{NCl}):

Moles of nitrobenzene (y)=0.01 mol\text{Moles of nitrobenzene } (y) = 0.01\text{ mol}

Thus, the mass of nitrobenzene (yy) is:

y=Moles×Molar mass=0.01 mol×123 g mol−1=1.23 gy = \text{Moles} \times \text{Molar mass} = 0.01\text{ mol} \times 123\text{ g mol}^{-1} = 1.23\text{ g}

Final Answer: The value of yy is 1.23.

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