Question

Given `100*g` EACH of carbon disulfide, and chlorine gas, `0.423*mol` carbon tetrachloride results. What is the percentage yield?

Answer 1

I get `"% yield"-=90%`...........

Explanation:

We need a stoichiometric equation....

`"CS"_2(g) +3"Cl"_2(g) stackrel(MnCl_2"/"AlCl_3)rarr "CCl"_4(g) +S_2"Cl_2(g)`

Note that this reaction SHOULD have been quoted with the question......this is what you are missing.

This is not a reaction I would be very happy to do....sulfide is oxidized to `stackrel(+I)S`, chlorine is reduced to `stackrel(-I)Cl`..and the reaction would pen and ink abominably.

`"Moles of carbon disulfide"=(100*g)/(76.13*g*mol^-1)=1.31*mol`

`"Moles of dichlorine"=(100*g)/(70.9*g*mol^-1)=1.41*mol`

Clearly, dichorine is the reagent in deficiency, and at most `(1.41*mol)/3=0.470*mol` with respect to carbon tet would result.....

We gots....`(65.0*g)/(153.82*g*mol^-1)=0.423*mol`

And so we gots a respectable yield of ......

`(0.423*mol)/(0.470*mol)xx100%=??%`