JEE Challenger
More from Amines
ChemHardAdv 2021 • Paper 2

Mass of Tin Required in Reduction of Nitrobenzene

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 xx.

Official Numerical Answer3.57

Topics & Concepts

AminesMole ConceptStoichiometryAminesOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the value of xx, we analyze the stoichiometric relations in the given sequence of reactions.

Step 1: Formula and Molar Mass of the Organic Salt

The reduction of nitrobenzene (C6H5NO2\text{C}_6\text{H}_5\text{NO}_2) in acidic medium (HCl\text{HCl}) yields anilinium chloride as the organic salt: Organic salt=Anilinium chloride (C6H5NH3+Cl−)\text{Organic salt} = \text{Anilinium chloride } (\text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^-)

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

Given mass of the organic salt = 1.29 g1.29 \text{ g}. Moles of organic salt=1.29 g129 g mol−1=0.01 mol\text{Moles of organic salt} = \frac{1.29 \text{ g}}{129 \text{ g mol}^{-1}} = 0.01 \text{ mol}

Step 2: Reaction of Sn\text{Sn} with HCl\text{HCl}

Tin (Sn\text{Sn}) reacts with HCl\text{HCl} to produce stannous chloride (SnCl2\text{SnCl}_2): Sn+2HCl→SnCl2+H2↑\text{Sn} + 2\text{HCl} \rightarrow \text{SnCl}_2 + \text{H}_2 \uparrow

From the balanced equation: 1 mole of Sn produces 1 mole of SnCl21 \text{ mole of } \text{Sn} \text{ produces } 1 \text{ mole of } \text{SnCl}_2

Step 3: Reduction of Nitrobenzene by SnCl2\text{SnCl}_2

The reaction of nitrobenzene with the salt (SnCl2\text{SnCl}_2) in the presence of HCl\text{HCl} is: C6H5NO2+3SnCl2+7HCl→C6H5NH3+Cl−+3SnCl4+2H2O\text{C}_6\text{H}_5\text{NO}_2 + 3\text{SnCl}_2 + 7\text{HCl} \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^- + 3\text{SnCl}_4 + 2\text{H}_2\text{O}

From the balanced equation: 3 moles of SnCl2 produce 1 mole of C6H5NH3+Cl−3 \text{ moles of } \text{SnCl}_2 \text{ produce } 1 \text{ mole of } \text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^-

Therefore, the moles of SnCl2\text{SnCl}_2 required to produce 0.01 mol0.01 \text{ mol} of the organic salt are: Moles of SnCl2=3×0.01=0.03 mol\text{Moles of } \text{SnCl}_2 = 3 \times 0.01 = 0.03 \text{ mol}

Step 4: Mass of Tin (xx) Required

Since 1 mole of Sn1 \text{ mole of } \text{Sn} gives 1 mole of SnCl21 \text{ mole of } \text{SnCl}_2: Moles of Sn=0.03 mol\text{Moles of } \text{Sn} = 0.03 \text{ mol}

Using the given molar mass of Sn=119 g mol−1\text{Sn} = 119 \text{ g mol}^{-1}: x=Moles of Sn×Molar mass of Snx = \text{Moles of } \text{Sn} \times \text{Molar mass of } \text{Sn} x=0.03×119=3.57 gx = 0.03 \times 119 = 3.57 \text{ g}

3.57

PreviousNext