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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

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superman1987

Answer:

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.

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