Question

Through multiple reactions, a chemist uses C, CaO, HCl, and H2O to produce dichlorobenzene, C4H4Cl2. Assuming that the efficiency of this lab is a minimum of 65.0%, how many molecules of dichlorobenzene could be produced in the lab from 0.500 kg of C?

(questions continued) ? The ratio of moles of C to C6H4Cl2 is determined to be 9:1.

I really do not understand this problem at all. I could use the help please.

Answer 1

`2.71 xx 10^24` molecules

Explanation:

With no limit on any of the reagents except the overall efficiency and amount of carbon, this would just be the number of molecules of dichlorobenzene - containing 6 atoms of C - that can be produced from 65% of the original carbon amount. IF multiple steps at 65% yield each are required, the number must be stated.

`0.65 xx 0.500 = 0.325"kg C"`, = `(325cancel"g")/(12cancel"g""/mol") = 27 "moles C"`
`27 "moles C" xx (1 "DCB")/(6C) = 4.50` moles Dichlorobenzene

`4.50 cancel"moles" xx 6.022 xx 10^23 "molecules"/cancel"mole" = 2.71 xx 10^24` molecules

The "ratio of moles of C to C6H4Cl2 as 9:1" is incorrect. It is obviously 6. It can't be "determined" any other way.