Moles of S203^2- = (molarity)(volume):

Question

=(0.032400 mol/L)(0.053 litres S203^2-)
= 0.03240 moles S203^2-

Moles if I2 based on Equation #2:
Moles I2 = (1/2)( 0.03240 moles S203^2-)
= .00086 moles I2

Moles of ClO^- based on Equation #1:
Moles ClO-(aq) = (2/3)(.00086 moles I2)
= .00057 moles ClO^3-

moles of ClO-(aq) = moles of NaOCl
.00057 moles ClO^3- = .00057 moles NaOCl

NaOCl mass = (molar mass * moles) = (74) (.00057) = 0.04235 g

Do we assume this gives us grams of NaOCl per 0.5 mL of bleach solution, since we didn’t divide or multiply 0.0005 L with anything? And to make it NaOCl / 100 mLs of bleach do we multiply the by 100?

Answer 1

Okay its not lettin me post up i dunt knwo why but the answer i got was

NaOCl mass = (molar mass * moles)
= (74) (.00057) = 0.04235 g

Do we assume this gives us grams of NaOCl per 0.5 mL of bleach solution, since we didn’t divide or multiply 0.0005 L with anything? And to make it NaOCl / 100 mLs of bleach do we multiply the by 100?

Answer 2

Moles of S203^2- = (molarity)(volume):
=(0.032400 mol/L)(0.053 litres S203^2-)
= 0.03240 moles S203^2-

I don't know if you made a typo or not but 0.03240 x 0.0530 is not 0.03240. You need to check your arithmetic. I get something like 0.064 or so for g NaOCl. That is g NaOCl in 0.5 mL (as GK said in his first response to your last posting). So you multiply this (again, just as he said) by 100/0.5 if you want the answer in g/100 mL. By the way, the molar mass of NaOCl is 74.44 and not 74.0. One reason you may be having trouble posting is you can not copy and paste. Try typing it in and you should have no trouble.