December 21, 2014

Homework Help: GK..please check...Chemistry Help Please

Posted by Sara on Wednesday, October 22, 2008 at 9:06pm.

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 didnít divide or multiply 0.0005 L with anything? And to make it NaOCl / 100 mLs of bleach do we multiply the by 100?

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