Posted by Sara on Sunday, October 12, 2008 at 5:55pm.
A 2.00 ml sample of an aqueous solution of hydrogen peroxide, H2O2 (aq) is treated with an excess of Kl (aq). The liberated I2 requires 12.40 mL of 0.1025 M Na2S2O3, for its titration. Is the H2O2 up to the full strength (3% H2O2 by mass) as an antiseptic solution? Assume the density of the aqueous solution of H2O2 ( aq) is 1.00 g/ mL.
H2O2(aq) + H + I^(aq) > H20(l)+I2 (not balanced)
S203^2 + I2 > S406 ^2 (aq) + I^ (aq) (not balanced)

Chemistryy....PLEASE HELPP HAVE MIDTERMMM  DrBob222, Sunday, October 12, 2008 at 6:18pm
H2O2 + 2H^+ + 2I^ ==> I2 + 2H2O check me out.
2S2O3^2 + I2 ==> 2I^ + S4O6^2
check me out.
Convert 12.40 mL x 0.1025 M S2O3^ to mols.
Using the coefficients in the balanced equations, convert mols S2O3^2 to mols I2 and from there to mols H2O2.
Convert mols H2O2 to grams.
Then calculate percent H2O2.
%H2O2 = (mass H2O2/mass solution)*100 = 
Chemistryy....PLEASE HELPP HAVE MIDTERMMM  Damon, Sunday, October 12, 2008 at 6:40pm
How many mol of S2O3^2?
12.4*10^3 L * .1025 M/L = 1.27*10^3 M of S2O3^2
balance
2 S2O3^2 + I2 = 2I^1 + S4 ...
so
one mol of S2O3 for every mol I (not I2)
so
1.27*10^3 mol of I
so
1.27*10^3 mol of KI
so
balance
H2O2+ 2H^1+ 2I^1 = 2H2O + I2
so one mol H2O2 for every 2 mol of I
so
.6355 *10^3 mol of H2O2 in 2*10^3 L
or
.317 mol of H2O2 in a Liter of solution
what is the mass of that?
.3178(2+32) = 10.8 grams/liter =.0108 g/mL
It is supposed to be .03 g/mL
so it is weak
check my arithmetic! 
Chemistryy....PLEASE HELPP HAVE MIDTERMMM  Sara, Monday, October 13, 2008 at 6:21am
Thank YOu soo mUch