Everything is ok except for the last step. The question is for % ClO2 and you substituted mass Cl2 and not ClO2.
That should be (25.9/26.98)*100 = ? and round to 3 significant figures.
Here is a short cut to calculating mols ClO2. You used ratio/proportion and that will work every time but you can also use the numbers in the balanced equation.
0.2 mol Cl2 x (2 mols ClO2/1 mol Cl2) = 0.4 mol ClO2. That's easy enough to do in your head whereas if I set it up with ratio/proportion it is harder to do in my head.
Thanks for showing your work. It helps us help you.
ClO2 is sometimes used as a chlorinating agent for water treatment.
Cl2 + 4 NaClO -> 4 NaCl + 2 ClO2
In an experiment, 1L Cl2 measured at 10C and 4.66atm is dissolved in .75L of 2.00M NaClO. If 25.9g of pure ClO2 is obtained, then what is the percent yeild for this experiment?
This is what I got:
I calculated the moles of Cl2 using PV=nrt, and got n= .200. For mass, I got Cl2= 14.18g
Then to get the moles of ClO2, I did a ration, .200/x = 1/2, x= .40 moles ClO2
Mass ClO2 = .4 X 67.45g/mol = 26.98g
Percent yield= 14.18g/26.98g x 100 = 52.56
Please tell me if there are any mistakes. Thank you!
1 answer