CS2 + 3Cl2 = CCl4 + S2Cl2
mol wt CS2 = 76
mol wt CCl4 = 154
67.1g CS2 = 67.1/76 = .88 moles CS2
45.2g CCl4 = 45.2/154 = .29 moles CCl4
.29/.88 = 32.9%
Hmm. Either I blew it, or the book is wrong. What did you get?
Carbon disulfide (CS2) reacts with excess chlorine (Cl2) to produce carbon tetrachloride (CCl4) and disulfur dichloride (S2Cl2). If 67.1 g of CS2 yields 45.2 g of CCl4, what is the percent yield? (Hint, you must first write the balanced equation.)
67.36
I tried this a couple of times, but I keep getting the wrong answer. Any help is appreciated!
5 answers
the number i put was what i got. what you did makes sense though, and i just checked and you were right. thank you!
the molecular weight of CS2 is 76.13
so the C in 67.1 g is:
67.1 * (12.01 / 76.13)
the molecular weight of CCl4 is 153.81
so the C in 45.2 g is:
45.2 * (12.01 / 153.81)
divide the CCl4 carbon by the CS2 carbon to find the percent yield
so the C in 67.1 g is:
67.1 * (12.01 / 76.13)
the molecular weight of CCl4 is 153.81
so the C in 45.2 g is:
45.2 * (12.01 / 153.81)
divide the CCl4 carbon by the CS2 carbon to find the percent yield
I think you'd better go with Scott's answer. H just considers the C yield.
> head slap <
duh.
> head slap <
duh.
Mg3N2 + H2O --> NH3 + Mg(OH)2