Can someone please help me? I keep getting a very large percentage at the very e
ID: 910289 • Letter: C
Question
Can someone please help me? I keep getting a very large percentage at the very end, which I don't think is plausible. Help would be much appreciated!!! thank you! Here are the relevant reaction equations:
Cl2(aq) + 2NaOH(aq) --> NaClO(aq) + NaCl(aq) + H2O(l)
ClO^-(aq) + 2 I^-(aq) + H2O(l) --> I2(aq) + Cl^-(aq) + 2OH^-(aq)
I2(aq) + I^-(aq) --> I3^-(aq)
I3^- + 2 S2O3^2- --> 3 I^- + S4O6^2-
A dilute bleach solution that is 5.00% by volume is prepared by mixing a portion of liquid bleach with DI H2O ; i.e. 5.00 mL bleach per 100 mL of solution to make the 5.00% solution. A 25.00 mL sample of this bleach solution is analyzed according to the procedure described in the lab. It is found that 23.13 mL of a 0.166 M solution of Na2S2O3 is needed to reach the stoichiometric endpoint of the titration.
1. Calculate the number of moles of S2O3^2- used for the titration.
2.Calculate the number of moles of I2 that reacted with the S2O3^2- during the titration.
3. Calculate the number of moles of NaClO present in this sample dilute bleach.
4. How many grams of Cl2 are equivalent to the number of moles of NaClO calculated above? The formula weight of Cl2 is 70.90 g/mol.
5. the 25.00 mL sample you analyzed is actually the dilute bleach solution. How many grams of undiluted bleach solution are actually present in the sample? the density of liquid bleach is 1.084 g/mL.
6.Calculate the precent "available chlorine" in the liquid bleach.
Please answer with steps! I would really like to figure out what I am doing wrong! thanks again!!
Explanation / Answer
Solution :-
1. Calculate the number of moles of S2O3^2- used for the titration.
Solution :-
Moles = molarity * volume
Moles of S2O3^2- = 0.166 mol per L * 0.02313 L = 0.002683 mol S2O3^2-
2.Calculate the number of moles of I2 that reacted with the S2O3^2- during the titration.
Solution :- mole ratio of the I2 to S2O3^2- is 1:2
Therefore
0.002683 mol S2O3^2- *1 mol I 2 / 2 mol S2O3^2- = 0.001342 mol I2
3. Calculate the number of moles of NaClO present in this sample dilute bleach.
Solution :-
1 mol I 2 = 1 mol NaClO
Therefore moles of NaClO = 0.001342 mol I2 * 1 mol NaClO / 1 mol I2 = 0.001342 mol NaClO
4. How many grams of Cl2 are equivalent to the number of moles of NaClO calculated above?
The formula weight of Cl2 is 70.90 g/mol.
Solution :- moles of Cl present = 0.001342 mol
So the moles of Cl2 equivalent = 0.001342 mol Cl * 1 mol Cl2 / 2 mol Cl = 0.000671 mol Cl2
Mass of Cl2 = 0.000671 mol * 70.90 g per mol = 0.0476 g Cl2
So the 0.0476 g Cl2 equivalent to the number of the molecules of the NaClO
5. the 25.00 mL sample you analyzed is actually the dilute bleach solution. How many grams of undiluted bleach solution are actually present in the sample? the density of liquid bleach is 1.084 g/mL.
Solution :-
Out of the 100 ml diluted solution 25 ml was used for the analysis
Therefore 100 ml solution contains 0.001342 mol * 74.44 g per mol NaClO = 0.10 g NaClO
So the mass of the NaClO in the 5 ml sample is 0.100 g
And density is 1.084 g per ml
So it contains 0.1 g NaClO
6.Calculate the precent "available chlorine" in the liquid bleach.
Solution :-
Mass of the Cl2 = 0.0476 g
Mass of solution = 5 ml * 1.084 g per ml = 5.42 g
% Cl = (0.0476 g / 5.42 g)*100%
= 0.878 % Cl