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
me = milliequivalents.
me IO3 = me NH2NH2 and
me = mL x N, then
mL x N = mL x N
Substitute and solve for mL IO3
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?
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.
To find the number of milliliters (mL) of 0.0257 N KIO3 needed to reach the end-point in the oxidation of 34.2 mL of 0.0416 N hydrazine, we need to use a balanced chemical equation and utilize the concept of stoichiometry.
First, let's write the balanced chemical equation for the oxidation reaction of hydrazine with KIO3:
5N2H4 + 2KIO3 + 13HCl → 6H2O + 2KCl + 5N2 + 8Cl2
The coefficients in the balanced equation tell us the molar ratio between the reactants and products. In this case, the ratio between hydrazine and KIO3 is 5:2.
To determine the number of moles of hydrazine (N2H4) in 34.2 mL of 0.0416 N solution, we can use the formula:
moles of solute = concentration (N) × volume (L)
Calculating the moles of hydrazine:
moles of N2H4 = 0.0416 N × 0.0342 L = 0.00142032 moles
Using the balanced equation, we know that the ratio of N2H4 to KIO3 is 5:2. Therefore, 5 moles of N2H4 react with 2 moles of KIO3.
moles of KIO3 = (0.00142032 moles N2H4) × (2 moles KIO3 / 5 moles N2H4) = 0.00056824 moles
To find the volume of 0.0257 N KIO3 needed, we can use the formula:
volume (L) = moles of solute / concentration (N)
Calculating the volume of 0.0257 N KIO3:
volume (L) = 0.00056824 moles / 0.0257 N = 0.02212 L
Since we want the volume in milliliters, we can convert the volume to mL by multiplying by 1000:
volume (mL) = 0.02212 L × 1000 = 22.12 mL
Therefore, approximately 22.12 mL of 0.0257 N KIO3 is needed to reach the end-point in the oxidation of 34.2 mL of 0.0416 N hydrazine in hydrochloric acid solution.