Question

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`?

Answer 1

`"% yield" = 88.8%`

Explanation:

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.

Start by writing the balanced chemical equation for the chlorination of ethane

`"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`

The reaction will always consume equal numbers of moles of each reactant, which means that chlorine gas will act as a limiting reagent here.

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.

Use chloroethane's molar mass to figure out how many grams would contain this many moles

`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"`

So, what does this tell you?

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

The difference between the theoretical yield of chloroethane, which is what you get when `100%` of the reactants are converted to chloroethane and hydrogen chloride, and the actual yield, which is what you get when some dichloroethane is produced, will represent the percent yield of chloroethane.

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

In your case, you will have

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