1. From the volume of titrant added, calculate the moles of vitamin C for each t
ID: 583595 • Letter: 1
Question
1. From the volume of titrant added, calculate the moles of vitamin C for each titration. Show a sample calculation for trial 1, and then just state the final results for the calculation of trials 2 and 3. Use 5 digits in all intermediate values and report the final value to the appropriate number of significant figures. 0.6612 g vitamin C KIO3- 0.002012 MM KI- 0.3 M HCl= 1.0 M Trial 1 Trial 3 40.62 14.00 26.62 Trial 2 26.49 Final buret reading37..21 Initial buret readin Volume of KIO3 10.00 27.21 26.49Explanation / Answer
Write the balanced equation of reaction vitamin C and KIO3
KIO3 + 3 C6H8O6 = 3 C6H6O6 + 3 H2O + KI
Concentration of KIO3 solution = 0.002012M (given)
For 1st trial
Volume of KIO3 added = 27.23 ml
Covert volume in L by dividing with 1000
27.23 ml/1000 = 0.02723 L
Calculate the moles of KIO3 by using formula n = c v
n = 0.002012M x (0.02723) L = 5.4789 x 10^-5 moles
Reaction stoichiometry show1:3 molar relationships between KIO3 and ascorbic acid.
So, moles of ascorbic acid = 3 moles of KIO3
= 3 x 5.4789 x 10^-5 moles
= 1.6437 x 10^-4 moles
= 1.64 x 10^-4 moles
For 2nd trial
Volume of KIO3 added = 26.49 ml
Calculate the moles of KIO3 by using formula n = c v
n = 0.002012M x (0.02649) L = 5.3298 x 10^-5 moles
Reaction stoichiometry show1:3 molar relationships between KIO3 and ascorbic acid.
So, moles of ascorbic acid = 3 moles of KIO3
= 3 x 5.3298 x 10^-5 moles
=1.5989 x 10^-4 moles
= 1.6 x 10^-4 moles
For 3rd trial
Volume of KIO3 added = 26.62 ml
Calculate the moles of KIO3 by using formula n = c v
n = 0.002012M x (0.02662) L = 5.3559 x 10^-5 moles
Reaction stoichiometry show1:3 molar relationships between KIO3 and ascorbic acid.
So, moles of ascorbic acid = 3 moles of KIO3
= 3 x 5.3559 x 10^-5 moles
=1.6068 x 10^-4 moles
= 1.61 x 10^-4 moles
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