The key here is to focus solely on the reaction that produces chloroethane,
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
So, you're mixing
$125 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "4.157 moles C"_2"H"_6$
$255color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.91color(red)(cancel(color(black)("g")))) = "3.596 moles Cl"_2$
The reaction will always consume equal numbers of moles of each reactant, which means that chlorine gas will act as a here.
More specifically, out of the
Now, you also have a
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
The difference between the theoretical yield of chloroethane, which is what you get when
$color(blue)("% yield" = "what you actually get"/"what you should get" xx 100)$
In your case, you will have
$"% yield" = (206 color(red)(cancel(color(black)("g"))))/(231.98color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.8 %")$