Moles of S203^2- = (molarity)(volume):
=(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?
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?
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.
To answer your question, let's break down the calculations and understand each step.
First, we have the moles of S203^2-:
Moles of S203^2- = (molarity)(volume) = (0.032400 mol/L)(0.053 L S203^2-) = 0.03240 moles S203^2-
Next, we have the moles of I2 based on Equation #2:
Moles I2 = (1/2)(0.03240 moles S203^2-) = 0.00086 moles I2
Then, we have the moles of ClO^- based on Equation #1:
Moles ClO-(aq) = (2/3)(0.00086 moles I2) = 0.00057 moles ClO^3-
Now, you are asking how to convert moles of ClO-(aq) to grams of NaOCl in a given amount of bleach solution.
Since the molar mass of NaOCl is 74 g/mol, we can calculate the mass of NaOCl using the formula: mass = (molar mass) * (moles).
Therefore, the mass of NaOCl in grams would be:
NaOCl mass = (74 g/mol) * (0.00057 moles) = 0.04235 g
Now, if you want to determine the grams of NaOCl per 0.5 mL of bleach solution, you need to divide the mass of NaOCl by the volume of bleach solution, which is 0.5 mL.
To convert this to NaOCl per 100 mL of bleach solution, you would then multiply the result by 100, since 0.5 mL is 1/200th of 100 mL.
I hope this clarifies the calculations for you. Let me know if you have any further questions!