Please check my answers.
1. Using the equations,
IO3- +5I- + 6H+ -> 3I2 + 3H2O
C6H8O6 + I2 -> C6H6O6 + 2I- + 2H+
how many equivalents of IO3- are required to produce enough I2 to oxidize one equivalent of ascorbic acid?
Answer: 1IO3- <-> 3I2 <-> 3C6H8O6
1/3 equivalents of IO3-
2. A 10 g sample of no name grape drink mix is diluted to 250 mL. 50 mL of grape drink is titrated with 35 mL of 0.005 M KIO3 solution.
a) How many moles of ascorbic acid are present in 50 mL of the grape solution?
0.005 mol/L KIO3 x 0.035 L KIO3 =
1.75 x 10^-4 mol KIO3
1.75 x 10^-4 mol KIO3 x (3 mol I2/1 mol KIO3) = 5.25 x 10^-4 mol I2.
3 KIO3 <-> 3 C6H8O6, so same amount of moles of C6H8O6.
b) What is the percent by mass of ascorbic acid in a package of grape drink mix?
5.25 x 10^-4 mol C6H8O6 x 176.124 g/mol C6H8O6 = 0.092 g C6H8O6
But this is grams in 50 mL of solution - we have 250 mL throughout which the ascorbic acid is distributed. So 0.092 g x 5 = 0.462 g C6H8O6 within 250 mL solution.
0.462 g/10.00 g x 100 = 4.62%
c) If there are 8 servings per package, calculate the %RDA of vitamin C in each serving of no name grape mix. Formula: (mg ascorbic acid/serving)/75 mg
(462 mg C6H8O6/8)/75 mg
= 0.77 % RDA
Thanks for your help!
I think all of your work is good except for 1. I believe the answer is 1/6 mol IO3^- = 1 equivalent ascorbic acid.
1 mole IO3^- = 3moles I2
3 moles ascorbic acid = 3 moles I2; therefore, 1 mole IO3^- = 3 moles I2
In the next step,
C6H8O6 + I2 ==> C6H6O2 +2H^+ + 2I^-
So this step is 1 mol C6H8O = 1 mol I2 but
2 equivalents C6H8O6 = 2 equivalents I2 (since I2 goes from 0 to -2 for 1 mole, that is 2e change for 1 mol, the equivalent weight I2 is 1/2 molar mass pr 2 equivalents. Therefore,
1/3 mol IO3^- = 1 mole I2 = 2 equivalents I2 and
1/6 mol IO3^- = 1 equivalent I2.
1/6 mol IO3^- = 1 equivalent ascprbic acid. You can check that out in your percent, which you did with moles and the 4.62% is correct.
If we do that with equivalents (and milliequivalents) we should get the same answer. m.e. = # milliequivalents and m.e.w. = milliequivalent weight.
So mL = 35 mL
Normality = 0.005*6 (since there are 6 equivalents in 1 mole IO3^-).
m.e.w. ascorbic acid = molar mass/2 = 176.124/2000 =0.088062
(mL x N x m.e.w./mass sample)*100 =
35.0mL x 0.03N x 0.088062g/2g sample)*100 = 4.62%
Viola!
Which of the following equations are equivalent to -2m - 5m - 8 = 3 + (-7) + m?
Your answers look correct! Here is a summary of the calculations:
a) The number of moles of KIO3 used in the titration is 1.75 x 10^-4 mol.
Using the stoichiometry from the equation, you found that the same number of moles of ascorbic acid (C6H8O6) are present in the grape solution as in the KIO3 solution, which is 5.25 x 10^-4 mol.
Therefore, there are 5.25 x 10^-4 mol of ascorbic acid in 50 mL of the grape solution.
b) You calculated the mass of ascorbic acid in 50 mL of the solution to be 0.092 g.
However, since the final result should be expressed as a percentage by mass in the package of grape drink mix, you multiplied the mass by 5 (since the dilution is by a factor of 5).
The mass of ascorbic acid in the package of grape drink mix is 0.462 g.
Finally, you calculated the percentage by mass of ascorbic acid in the package, which is 4.62%.
c) You correctly used the mass of ascorbic acid in the package (0.462 g) and divided it by the number of servings (8) to find the mass per serving (0.05775 g).
Since the recommended daily allowance (RDA) of vitamin C is given as 75 mg, you converted the mass per serving to milligrams (57.75 mg).
Finally, you calculated the percentage of the RDA of vitamin C in each serving, which is 0.77%.
Overall, your calculations and answers appear to be correct. Well done!
1. To determine how many equivalents of IO3- are required to produce enough I2 to oxidize one equivalent of ascorbic acid, we look at the stoichiometric ratios in the given equations. From the first equation, we can see that 1 equivalent of IO3- produces 3 equivalents of I2. From the second equation, we can see that 1 equivalent of I2 is required to oxidize 1 equivalent of ascorbic acid.
Therefore, the overall stoichiometry is 1 IO3- : 3 I2 : 1 C6H8O6. This means that for every 1 equivalent of ascorbic acid, we need 1/3 equivalents of IO3-.
So the answer is 1/3 equivalents of IO3- are required to produce enough I2 to oxidize one equivalent of ascorbic acid.
2a. To determine the number of moles of ascorbic acid present in 50 mL of the grape solution, we need to use the given concentration of KIO3 solution and the volume of KIO3 solution used in the titration.
The number of moles of KIO3 can be calculated using the equation:
moles of KIO3 = concentration of KIO3 solution (mol/L) × volume of KIO3 solution (L)
So, 0.005 mol/L × 0.035 L = 1.75 × 10^-4 mol KIO3
Since the stoichiometry between KIO3 and ascorbic acid is 1:3, the number of moles of ascorbic acid is also 1.75 × 10^-4 mol.
2b. To calculate the percent by mass of ascorbic acid in a package of grape drink mix, we first need to find the mass of ascorbic acid in the 50 mL of the grape solution.
Using the stoichiometry from part 2a, we found that there are 1.75 × 10^-4 mol of ascorbic acid in 50 mL of the solution.
Next, we can use the molar mass of ascorbic acid (C6H8O6) to calculate the mass:
mass of ascorbic acid = moles of ascorbic acid × molar mass of ascorbic acid
mass of ascorbic acid = 1.75 × 10^-4 mol × 176.124 g/mol = 0.0308 g
However, this calculation gives the mass in 50 mL of the solution, but we need the mass in the total volume of the solution (250 mL). To find the mass in the 250 mL solution, we multiply by a factor of 5:
mass of ascorbic acid in 250 mL = 0.0308 g × 5 = 0.154 g
Finally, we can calculate the percent by mass:
percent by mass of ascorbic acid = (mass of ascorbic acid in 250 mL / total mass of grape drink mix) × 100
percent by mass of ascorbic acid = (0.154 g / 10.00 g) × 100 = 1.54%
So, the percent by mass of ascorbic acid in a package of grape drink mix is 1.54%.
2c. To calculate the percent Recommended Daily Allowance (RDA) of vitamin C in each serving of the grape mix, we need to know the RDA value and the amount of ascorbic acid per serving.
Given that there are 8 servings per package and the total amount of ascorbic acid in the package is 0.154 g (from part 2b), we can calculate the amount of ascorbic acid per serving:
amount of ascorbic acid per serving = (0.154 g / 8 servings) = 0.01925 g
Assuming the RDA for vitamin C is 75 mg, we can convert grams to milligrams:
0.01925 g = 19.25 mg
Finally, we can calculate the percent RDA:
percent RDA = (amount of ascorbic acid per serving / RDA value) × 100
percent RDA = (19.25 mg / 75 mg) × 100 = 25.67%
So, the percent RDA of vitamin C in each serving of the grape mix is 25.67%.