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) –> 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 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 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?

= .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?

To calculate the grams of NaOCl per 0.5 mL of bleach solution, you need to convert the number of moles of NaOCl to grams by multiplying it by the molar mass of NaOCl (which is 74 g/mol). Therefore, the equation would be:

Grams of NaOCl per 0.5 mL = (moles of NaOCl) * (molar mass of NaOCl)

In the given example, we have already calculated the moles of NaOCl to be 0.00057 moles. Therefore, the calculation would be:

Grams of NaOCl per 0.5 mL = (0.00057 moles) * (74 g/mol) = 0.04235 g

Now, if you want to calculate the grams of NaOCl per 100 mL of bleach solution, you need to multiply the above result by 100. The reason we multiply by 100 is because we are scaling up the volume of the bleach solution from 0.5 mL to 100 mL. Therefore, the equation would be:

Grams of NaOCl per 100 mL = (grams of NaOCl per 0.5 mL) * 100

Using the previously calculated value of 0.04235 g for grams of NaOCl per 0.5 mL, the calculation would be:

Grams of NaOCl per 100 mL = (0.04235 g) * 100 = 4.235 g

Hence, the grams of NaOCl per 100 mL of bleach solution is 4.235 g.