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.
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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
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.
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
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!
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
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.
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.
To get the moles of S2O3^2-, we can use the formula: moles = molarity x volume.
Trail 1:
Molarity of S2O3^2- = 0.03240 mol/L
Volume of S2O3^2- = 0.053 L
moles of S2O3^2- = (0.03240 mol/L) x (0.053 L S2O3^2-)
= 0.00172 moles S2O3^2-
To get the moles of I2, we can use the stoichiometry of the reaction:
2S2O3^2- + I2 -> 2I- + S4O6^2-
Since the ratio of S2O3^2- to I2 is 2:1, we can divide the moles of S2O3^2- by 2 to get the moles of I2.
moles of I2 = (1/2)(0.00172) = 0.00086 moles
To get the moles of ClO3-, we can use the stoichiometry of the reaction:
2ClO- + 4H+ + 6I- -> 3I2 + 2Cl- + 2H2O
Since the ratio of I2 to ClO3- is 3:2, we can multiply the moles of I2 by (2/3) to get the moles of ClO3-.
moles of ClO3- = (2/3)(0.00086) = 0.00057 moles
Since the reaction specifies that the moles of ClO3- is equal to the moles of NaOCl, we can use the molar mass of NaOCl to calculate the mass of NaOCl.
Molar mass of NaOCl = 74.44 g/mol
mass of NaOCl = (74.44 g/mol)(0.00057 moles)
= 0.0424 g
To convert the mass of NaOCl from 0.5 mL to 100 mL, we can use the equation: (mass)(100 mL/0.5 mL).
mass of NaOCl over 100 mL = (0.0424 g)(100 mL/0.5 mL)
= 8.4861 g/100 mL
For trail 2, you can repeat the same calculation steps using the given values for volume and molarity of S2O3^2-.
moles of NaOCl for trail 2 = 0.000598 moles
mass of NaOCl for trail 2 = (74.44 g/mol)(0.000598 moles)
= 0.0445 g
To convert the mass of NaOCl for trail 2 from 0.5 mL to 100 mL, we can use the equation: (mass)(100 mL/0.5 mL).
mass of NaOCl over 100 mL for trail 2 = (0.0445 g)(100 mL/0.5 mL)
= 8.908 g/100 mL