Ask Question
11 August, 08:31

What is the theoretical yield of ethyl chloride in the reaction of 27.17 g of ethylene with 58.33 g of hydrogen chloride? (For ethylene, MW=28.0 amu; for hydrogen chloride, MW=36.5 amu; for ethyl chloride, MW=64.5 amu.) Report your answer to the nearest hundredth of a gram without units.

H2C=CH2 + HCl → CH3CH2Cl

+4
Answers (1)
  1. 11 August, 08:49
    0
    63.83.

    Explanation:

    For the reaction: H₂C=CH₂ + HCl → CH₃CH₂Cl.

    1.0 mole of ethylene reacts with 1.0 mol of HCl to produce ethyl chloride.

    We need to calculate the no. of moles of ethylene chloride and HCl:

    no. of moles of ethylene = mass/molar mass = (27.71 g) / (28.0 g/mol) = 0.9896 mol.

    no. of moles of HCl = mass/molar mass = (58.33 g) / (36.5 g/mol) = 1.589 mol.

    Since, from the balanced equation; ethylene reacts with HCl with 1: 1 ratio. So, ethylene is the limiting reactant and HCl is in excess.

    Now, we need to find the no. of moles of produced ethyl chloride:

    ∵ 1.0 mole of ethylene produce → 1.0 mole of CH₃CH₂Cl.

    ∴ 0.9896 mole of ethylene produce → 0.9896 mole of CH₃CH₂Cl.

    The mass produces of ethyl chloride = no. of moles x molar mass = (0.9896 mol) (64.5 g/mol) = 63.83 g.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “What is the theoretical yield of ethyl chloride in the reaction of 27.17 g of ethylene with 58.33 g of hydrogen chloride? (For ethylene, ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers