GK or Dr.Bob222...Chemistry please help...

Please refer to the previous post Chemistry Help Please.

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

Posted by DrBob222 on Wednesday, October 22, 2008 at 9:37pm in response to GK..please check...Chemistry Help Please

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.

__________________________________________

Yes, i had made a typo, i am posting my calculation below, please double check..were i am going wrong...since the answer i am getting is different from what you got .

MOLARITY OF S2O3^2- = 0.03240 mol/L

TRAIL 1: VOLUME OF S2O3^2- = 0.053L
TRAIL 2: VOLUME OF S2O3^2- = 0.0554L

Reaction 1:

2ClO- (aq)+ 4H^+(aq)+ 6I-(aq) ---> 3I2(aq)+ 2Cl-(aq) + 2H2O

Reaction 2:

2S203^2-(aq)+ I2 ----> 2I-(aq)+ S4O6^2- (aq)]

TRAIL 1:

moles of S2O3^2- : (molarity)(volume)
= (0.03240 mol/L)(0.053 L S203^2-)
= 0.00172 moles S203^2-

MOLES OF I2:
=(1/2)(0.00172)
=.00086 MOLES

MOLES OF ClO3- :
= (2/3)(.00086)
= .00057

MOLES OF ClO3- = MOLES OF NaOCl

Molar mass of NaOCl - 74.44

mass of NaOCl= (74.44)(0.00057 moles)
= 0.0424 g

TO get NaOCl over 100 ml= (0.0424g)(100mL/0.5mL)= 8.4861 g/mL

So this is for trail 1, and for trail 2 after doing all the calculation i got :

moles of NaOCl= 0.000598
mass = (74.44)(0.000598)
= 0.0445

(0.0445)(100/0.5) = 8.908

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asked by Sara
  1. I don't have that equation. Is yours balanced? I don't think so. The atoms balance but the charge does not. You have -4 charge on the left and -2 on the right side. In addition the loss and gain of electrons is not the same. I tried to find the initial posting where GK wrote the equations for you BECAUSE he posted a sequel showing that the 2/3 was not correct. I couldn't locate it.
    So,
    1. You need to redo the equation of
    ClO^- + 2I^- ==> I2 + Cl^-
    The titration with S2O3^-2 is ok.
    2. You are rounding 0.00086 and that is not very good. Since you have 4 significant figures in the molarity, you should have 4 in the mols so instead of 0.00086 it actually should be 0.0008586.
    If you correct that (which is ok to that point) and re-establish the stoichiometry for the equations, you should be ok.

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  2. tHIS IS WHAT GK POSTED:

    #

    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)


    # Chemistry Help Please - GK, Wednesday, October 22, 2008 at 11:32am

    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-)

    # Chemistry Help Please - GK, Wednesday, October 22, 2008 at 11:48am

    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

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    posted by Saira
  3. I think the equation posted for equation #1 is incorrect. You balance it and see what you get. However, you can see, can't you, that it isn't balanced because the charges don't balance on both sides of the equation.
    I have
    ClO^- + 2I^- + 2H^+ ==>H2O + I2 + Cl^-
    1 Cl on each side.
    1 O on each side.
    2 I on each side.
    2 H on each side.
    -1 charge on left and right.
    electron change: +1 on Cl to -1 on right is gain of two electrons. -1 on I makes -2 total on left and 0 on the right which is loss of 2 electrons.
    atoms balance. charge balances. electron change balances. voila!

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  4. SO it for every 1 mol of ClO3- there is 1 mol of I2 :

    Therefore, even the NaClO is 0.0008586

    multiplied by molar mass is 0.06391

    and then multilplied by (100/.05) is 12.78

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    posted by Saira
  5. I don't know where you get ClO3^- but everything else looks ok if I remember the problem correctly. 12.7% sounds high to me; most of the hypochlorite I see on the grocery shelf is 4-6%; of course, I think I remember the problem saying that you took 0.5 mL of a prepared solution. I suppose the person making the solution could have made it any concentration s/he wished.

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  6. 12.7 is the mass of NaOCL/100 mL bleach ..omg just ignore the 3 i was working on a different question which has KIO3- so mixing every up.

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    posted by Saira

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