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Question 59

The mass of NH$$_3$$ produced when $$131.8$$ kg of cyclohexane carbaldehyde undergoes Tollen's test is _____ kg. (Nearest Integer)
Molar mass of C $$= 12$$ g/mol, N $$= 14$$ g/mol, O $$= 16$$ g/mol


Correct Answer: 80

Solution

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Step 1: Molar Mass Calculations

When an aldehyde undergoes oxidation with Tollen's reagent, the balanced chemical equation is:

$$R-CHO+2[Ag(NH_3)_2]^++3OH^-→R-COO^-+2Ag↓+4NH_3+2H_2O$$

From the stoichiometry of this reaction, 1 mole of aldehyde produces exactly 4 moles of ammonia (NH3)

  • Cyclohexane carbaldehyde ($$C_{7}H_{12}O$$):
    Molar Mass $$=(7\times12)+(12\times1)+(1\times16)=84+12+16=112g/mol$$
  • Ammonia ($$NH_3$$):
    Molar Mass $$= 14 + (3 × 1) = 17 g/mol$$
Step 2: Number of Moles of Reactant

Given mass of cyclohexane carbaldehyde $$= 131.8 kg = 131800 g$$

Moles of aldehyde = Given Mass / Molar Mass $$= 131800 g / 112 g/mol ≈ 1176.78 moles$$
Step 3: Moles and Mass of $$NH_3$$ Produced

According to the 1 : 4 stoichiometric ratio:

Moles of $$NH_3$$ = 4 × Moles of aldehyde $$= 4 × 1176.78 ≈ 4707.14 moles$$
Mass of $$NH_3$$ = Moles × Molar Mass of $$NH_3$$ $$= 4707.14 moles × 17 g/mol ≈ 80021.4 g ≈ 80 kg$$

Correct Numerical Answer: 80

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