Calculating the percent yield...?

Question

When ethane, C2H6, reacts with chlorine, Cl2, the main product is C2H5Cl, but other products containing Cl such as C2H4Cl2,are also obtained in small quantities. The formation of these other products reduces the yield of C2H5Cl. Calculate the percent yield of C2H5Cl if the reaction of 125 g of C2H6 with 255 g of Cl2 produces g of C2H5Cl

Thank you in advance for the help

10 points

Answer 1

molar masses:

C2H6  ; 30.07 g/mol

C2H5Cl : 64.51 g /mol

find how much could have been produced:

125 g C2H6 @ 64.51g C2H5Cl  / 30.07 g C2H6 = 268.17 grams

Calculating the percent yield:

255 g / 268g times 100 = 95.09  % yield

your answer (3sig figs) : 95.1%

Answer 2

percent yield = (actual yield / theoretical yield)*100%

reaction equation 2 C2H6 + Cl2 ----> 2 C2H5Cl + what ever

Okay we have to find the limiting agent

moalr mass ethane = 30.07 g/mol http://en.wikipedia.org/wiki/Ethane

molar mass chlorine = 35.453*2= 70.906 http://www.elementsdatabase.com/Chlorine...

so # moles = mass/molar mass

for ethane  125/30.07 = 4.15696 moles

for chlorine  255/70.906 = 3.5963 moles

from the reaction equation and the number of moles we can see that the ethane is the limiting reagent so for the theoretical yield we will use it because there is an excess of Cl2.

from the reaction equation 2 moles C2H6 = 2 moles C2H5Cl

the molar ratio is 1:1

so 4.15696 moles C2H6 would produce 4.15696 moles C2H5Cl at 100% reaction

the molar mass of C2H5Cl is 64.51 g/mol http://en.wikipedia.org/wiki/Chloroethan...

4.15696 moles*64.51 g/mole = 268.16 grams of C2H5Cl

268.16 grams of C2H5Cl this is the theoretical yield

So  the percent yield = (actual yield / theoretical yield)*100%

percent yield = (actual yield / 268.16 grams of C2H5Cl)*100%

XXXXXXXXXXXXXXXXXXXXXXXXXXX I need the mass of the C2H5Cl produced to finish the equation. The amount actually produced would be the actual yield. Thank you. Steve calculation above is assuming that you produced 255 g of C2H5Cl. You left the space empty so we don't know how much C2H5Cl  was produced and with out it we can't finish the equation. Ever.