Question:

The amount of ethane produced in the following reaction is kg.
C2H4 (2 kg) + H2 (2 kg) Wilkinson ′s Catalyst C2H6 (90% catalytic conversion)
(round off to two decimal places)

Updated On: Nov 14, 2025
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Correct Answer: 1.92 - 1.93

Solution and Explanation

The balanced chemical reaction is:

C2H4 + H2 → C2H6 (90% catalytic conversion)

Step 1: Determine the moles of reactants

  • Molar mass of C2H4 = 28 g/mol
  • Molar mass of H2 = 2 g/mol
  • Moles of C2H4 = (2000 g)/(28 g/mol) ≈ 71.43 mol
  • Moles of H2 = (2000 g)/(2 g/mol) = 1000 mol

 

Step 2: Identify the limiting reactant

  • For full conversion, 71.43 mol of C2H4 requires 71.43 mol of H2.
  • C2H4 is the limiting reactant since all of it reacts first.

 

Step 3: Calculate the moles of C2H6 produced

  • Moles of C2H6 = 71.43 mol (90% conversion) = 71.43 * 0.90 ≈ 64.29 mol

 

Step 4: Calculate the mass of C2H6 produced

  • Molar mass of C2H6 = 30 g/mol
  • Mass of C2H6 = 64.29 mol * 30 g/mol = 1928.57 g = 1.93 kg

 

The mass of ethane produced is 1.93 kg.

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