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?

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.

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)

You should multilply .00469 by 3, as for each mole of KIO3 you get 3 moles of I3

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.

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.

To determine the number of grams of I3- produced, we first need to calculate the number of moles of each reactant and then compare the stoichiometric ratio in the balanced equation.

1. Find the number of moles of potassium iodate (IO3-):
Moles of IO3- = mass / molar mass
The molar mass of IO3- is 1 x (atomic mass of I) + 3 x (atomic mass of O)
The atomic mass of I is 126.90 g/mol
The atomic mass of O is 16.00 g/mol
Molar mass of IO3- = 126.90 + 3(16.00) = 166.90 g/mol
Moles of IO3- = 1.005 g / 166.90 g/mol

2. Find the number of moles of potassium iodide (KI):
Moles of KI = mass / molar mass
The molar mass of KI is the sum of the atomic masses of K and I
The atomic mass of K is 39.10 g/mol
The atomic mass of I is 126.90 g/mol
Molar mass of KI = 39.10 + 126.90 = 166.00 g/mol
Moles of KI = 0.5 g / 166.00 g/mol

3. Determine the limiting reagent:
Compare the moles of reactants using the stoichiometric ratio from the balanced equation.
According to the balanced equation, the mole ratio of IO3- to KI is 1:8.
Therefore, for every 1 mole of IO3-, you need 8 moles of KI.
If the moles of KI are less than what is required by the ratio, KI is the limiting reagent. Otherwise, IO3- is the limiting reagent.

4. Once you have identified the limiting reagent, use its moles to calculate the moles of I3- produced:
The mole ratio of IO3- to I3- is 1:3 in the balanced equation.
Multiply the moles of the limiting reagent by the mole ratio to find the moles of I3- produced.

5. Finally, calculate the grams of I3- produced:
Grams of I3- = moles of I3- x molar mass of I3-

Please provide the moles of KI to proceed with the calculation.