When ethane, #C_2H_6#, reacts with chlorine gas the main product is #C_2H_5Cl# but small amounts of #C_2H_4Cl_2# are produced. What is the percent yield of #C_2H_5Cl# if the reaction of 125g of ethane with 255g of chlorine gas produced 206g of #C2H_5Cl#?

Question

Ellie Andrews · Accepted Answer

#"% yield" = 88.8%#

The key here is to focus solely on the reaction that produces chloroethane, #"C"_2"H"_5"Cl"#, and ignore the one that produces dichloroethane, #"C"_2"H"_4"Cl"#, the side product of the reaction.

The balanced chemical equation for the chlorination of ethane should be written first.

#"C"_2"H"_text(6(g]) + "Cl"_text(2(g]) -> "C"_2"H"_5"Cl"_text((g]) + "HCl"_text((g])#

Your next step will be to use the #1:1# mole ratio that exists between ethane and chlorine gas to determine whether or not you're dealing with a limiting reagent.

So, you're mixing #"125 g"# of ethane with #"255 g"# of chlorine gas. Convert the masses of the two reactants to moles by using their respective molar masses

#125 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "4.157 moles C"_2"H"_6#

#255color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.91color(red)(cancel(color(black)("g")))) = "3.596 moles Cl"_2#

Since the reaction will always use the same amount of moles of each reactant, the limiting reagent in this case will be chlorine gas.

More specifically, out of the #4.157# moles of ethane, only #3.596# will actually take part in the reaction. The rest will be in excess.

Now, you also have a #1:1# mole ratio between the reactants and chloroethane. This means that the reaction will produce #3.596# moles of chloroethane, since that's how many moles of each reactant take part in the reaction.

Determine the number of grams that would contain this many moles using the molar mass of chloroethane.

#3.596 color(red)(cancel(color(black)("moles C"_2"H"_5"Cl"))) * "64.51 g"/(1color(red)(cancel(color(black)("mole C"_2"H"_5"Cl")))) = "231.98 g"#

What does this indicate, then?

In theory, the reaction should produce #"231.98 g"# of chloroethane. However, you know that the reaction only produced #"206 g"# of chloroethane.

The percent yield of chloroethane is determined by comparing the theoretical yield, which is obtained when all reactants are converted to hydrogen chloride and chloroethane, with the actual yield, which is obtained when some dichloroethane is produced.

#color(blue)("% yield" = "what you actually get"/"what you should get" xx 100)#

In your situation, you'll have

#"% yield" = (206 color(red)(cancel(color(black)("g"))))/(231.98color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.8 %")#

Cooper Adams · Answer

undefined Percent yield of C₂H₅Cl is 66.2%.