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

63.83.

Explanation:

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

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.

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.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

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