been stuck for 4 hours currently... need help now... relevant reaction equations

Question

been stuck for 4 hours currently... need help now...

relevant reaction equations:

Cl2(aq) 2NaOH (aq) --> NaClO (aq) + H2O (l) + NaCl (aq)

ClO ^- (aq) + 2I ^- + H2O (l) --> I2 (aq) + Cl ^- (aq) + 2O ^- (aq)

I2 (aq) + I ^- (aq) --> I3 ^- (aq)

I3 ^- (aq) + 2S2O3 ^2- (aq) --> 3I ^- +S4O6 ^2- (aq)

1) a dilute bleach solution that is 5.00% by volume is prepared by mixing a portion of liquid bleach with DI water.

example: 5.00mL bleach per 100mL of solution to make the 5.00% solution. A 25.00mL sample of this bleach solution is analyzed according to the procedure described in the lab. it is found that 24.32 mL of a 0.167M solution of Na2S2O3 is needed to reach the stoichiometric endpoint of the titration. calculate the number of moles 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 of the dilute bleach solution.

4) How man 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 a dilte solution of liquid bleach. how many grams of undiluted liquid bleach are actually present in this sample? the density of liquid bleach is 1.084 g/ml.

6) calculate the percent "available chlorine" in the liquid bleach.

please show all the works. thanks

Explanation / Answer

Number of moles of Na2S2O3 = Number of moles of S2O3(2-) = 24.32/1000*(0.167) = 4.06144*10^(-3) moles

Number of moles of I2 = half the number of moles of Na2S2O3 = 2.03072 * 10^(-3) moles

Number of moles present = 5/1000 = 0.05*25/100 = 0.0125 moles

Grams of Cl2 = 0.0125 * 70.90 = 0.88625 gms

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