You have the equation.
Convert 1.005 g KIO3 to moles. #moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles KIO3 to moles I3^- (I'm sure you meant I3^- and not L3^-).
Now convert moles I3^- to grams.
grams = moles x molar mass.
1.005 grams of potassium iodate is reacted with .5 grams of potassium iodide. THe net ionic reaction is shown below.
IO3-+8I-+6H+-->3I3-+3H2O
How many grams of the L3- are produced?
5 answers
This is what I keep trying and am not coming up with the answer that he gave us (.43grams)
Homework:
KIO3- is 1.005g/214amu=.00469
.00469x1/3=.00156
.00156x380.7=.593 grams (I amu x 3=380.7)
Is there something I am doing wrong with my math or is the teachers answer of .43 incorrect? (he has been incorrect before on accident)
Homework:
KIO3- is 1.005g/214amu=.00469
.00469x1/3=.00156
.00156x380.7=.593 grams (I amu x 3=380.7)
Is there something I am doing wrong with my math or is the teachers answer of .43 incorrect? (he has been incorrect before on accident)
You should multilply .00469 by 3, as for each mole of KIO3 you get 3 moles of I3
moles I3=3x.00469
moles I3=3x.00469
I didn't even see the 0.5 KI until you reposted. That is the problem. This is a limiting reagent problem AND you made a chemical error also.
For the chemical error, which won't affect the way you must work the problem, is that you don't multiply by 1/3 to convert from moles KIO3 to moles I3^-. You multiply by 3; that is,
moles I3^- = moles KIO3 x (3 moles I3^-/1 mole KIO3) = ??
If you will run the moles KI in 0.500 gram, and check the ratio of KIO3 to KI, the KI is the limiting reagent; therefore,
mols KI = 0.5/molar mass.
moles I3^- = moles KI x (3/8) = ??
grams I3^- = moles I3^- x 380.71.
Watch the significant figures if your prof is a s.f. freak. I obtained 0.43 g when I worked it.
For the chemical error, which won't affect the way you must work the problem, is that you don't multiply by 1/3 to convert from moles KIO3 to moles I3^-. You multiply by 3; that is,
moles I3^- = moles KIO3 x (3 moles I3^-/1 mole KIO3) = ??
If you will run the moles KI in 0.500 gram, and check the ratio of KIO3 to KI, the KI is the limiting reagent; therefore,
mols KI = 0.5/molar mass.
moles I3^- = moles KI x (3/8) = ??
grams I3^- = moles I3^- x 380.71.
Watch the significant figures if your prof is a s.f. freak. I obtained 0.43 g when I worked it.
I really appreciate the help, I understand all the concepts I just get overwhelmed with all the steps sometimes, when I see all those numbers and letters, This just saved me hours of confusing myself even further if I did not have any help.