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

Posted by Sara on Wednesday, October 22, 2008 at 9:03pm.

Sorry not leting me post entire thing on last post..can you please check Thank- YOu

# Chemistry Help Please - GK, Wednesday, October 22, 2008 at 11:24am

Reaction #1 (properly balanced):
2ClO-(aq) + 4H+(aq) + 6I-(aq) Ė> 3I2 + 2Cl-(aq) + 2H2O

Reaction #2 (properly balanced):
2S203^2-(aq) + I2 ---- > 2I-(aq) + S4O6^2- (aq)

Moles of S203^2- = (molarity)(volume):
(0.032400 mol/L)(__liters S203^2-) = __moles S203^2-

Moles if I2 based on Equation #2:
Moles I2 = (2)(__moles S203^2-)

Moles of ClO^- based on Equation #1:
Moles ClO-(aq) = (2/3)(__moles I2)


Correction with apologies. The corrected mole ratio is in bold print:
Moles if I2 based on Equation #2:
Moles I2 = (1/2)(__moles S203^2-)

One more comment:
The final answer assumes that:
moles of ClO-(aq) = moles of NaClO
Based on that, convert the number of moles to grams of NaClO by multiplying the number of moles by the formula mass of NaClO. That would give you the grams of NaClO per 0.5 mL of bleach solution.
To get grams NaClO / 100 mLs of bleach, what would you do?

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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