How many mL of 0.0257 N KIO3 would be needed to reach the end-point in the oxidation of 34.2 mL of 0.0416N hydrazine in hydrochloric acid solution

Answer 1

me = milliequivalents.
me IO3 = me NH2NH2 and
me = mL x N, then
mL x N = mL x N
Substitute and solve for mL IO3

Answer 2

This exercise is from Schaum's book of chemistry, and the answer has to be 55,4 ml but I don't get that result ( my result is 51,7 ml).
I did all the steps and solve for mL of IO3 , is there something that I'm missing?
My steps were:
Vhidrazine×Chidrazine=VKIO3×CKIO3
VKIO3=Vhidrazine×Chidrazine/CKIO3
Is the book wrong?

Answer 3

No. Schaum's is right. If I do it I get 55.3587 mL which rounds to 55.4 mL. You must be punching in the wrong numbers or the wrong key somewhere. Make sure you haven't transposed a number.