Chemistry, 29.02.2020 02:54 kiarabermudez754

Calculate the weight percent of ascorbic acid in a tablet of Vitamin C from the following data:A 80 mg sample of a crushed Vitamin C tablet was dissolved in 40 mL of H2SO4 and 20 mL of water. Two grams of KI and 40. mL of 0.00653 M KIO3 solution was added, and the mixture titrated to a starch endpoint. The titration required 15 mL of 0.0510 M thiosulfate solution. I know the answer is 88%, but how did they get it?

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Answer from: alexandroperez13

The Answer is 88%

Explanation:

The balanced ionic equation of the reaction

$IO^{3- }+ 8 I^- + 6 H^+ = 3 I^{3- }+ 3 H_2O$

$I^{3-} + 2 S_2O_3^{2-} = 3 I^- + S_4O_6^{2-}$

$C_6H_8O_6 + I^{3- }+ H_2O = C_6H_6O_6 + 3 I^- + 2 H^+$

Looking at the above reactions

The original number of moles of $I^{3-}$ = 3 × number of moles of $IO^{3-}$

= 3 × volume × concentration of KIO_3

Note: The formula for number of moles is volume × concentration

$=3 * \frac{40}{1000} * 0.00653 =0.000784$ mol

The number of moles of $I^{3-}$ left after its reaction with ascorbic acid

= $\frac{1}{2}$ x moles of $S_2O_3^{2-}$

= $\frac{1}{2}$ x volume x concentration of $S_2O_3^{2-}$

$= \frac{1}{2} * \frac{15}{1000} * 0.0510 =0.000383 \ mol$

Note: The division by 1000 is to convert mill liter to liter

Moles of ascorbic acid = moles of $I^{3-}$ reacted

$= initial \ moles \ of \ I^{3-} \ - remaining \ moles \ of \ I^{3-}$

=0.000784 - 0.000383

$=0.000401 \ mols$

$Mass = moles \ * molar \ mass$

Hence

Mass of ascorbic acid = moles of ascorbic acid × molar mass of ascorbic acid

$= 0.000401* 176.12 = 0.07055\ g$

$= 70.55\ mg$

Weight% of ascorbic acid = mass of ascorbic acid/mass of sample x 100%

= 70.55/80 × 100%

= 88.1%

Answer from: Quest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

Answer from: Quest

Only one reactant yields 2 or more products. meaning you will only have 1 product and it will break down into 2 or more products. eg. h2o -> h2 + o2

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Calculate the weight percent of ascorbic acid in a tablet of Vitamin C from the following data:A 80...