Please please please answer this question fully and please answer it only if you

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Please please please answer this question fully and please answer it only if you're 100% of the correct answer! Thank you! Some people have an allergic reaction to the food preservative sulfite (so32 Sulfite in wine was measured by the following procedure: To 50.0 mL of wine were added 5.00 mL of solution containing (0.8043 g KIO3 5 g KI)/100 mL. Acidification with 1.0 mL of 6.0 M H2SO4 quantitatively converted IO3 i nto 13 The 13 reacted with SO3 to generate SO leaving excess 13 n solution. The excess I3 required 12.86 mL of 0.04818 M Na2s203 to reach a starch end point. (a) Write the reaction that occurs between KIO3 and KI when the H2So4 is added. (Use the lowest possible coefficients. omit spectator ions and states-of-matter.) chem Pad Help (b) Write a balanced reaction between 13 and sulfite. Use the lowest possible coefficients. Omit spectator ions and states-of-matter.) chem Pad Help (c) Find the concentration of sulfite in the wine. Express your answer in mol/L and in mg so32 per liter. mol/L HINTS: Don't use scientific notation, do put zero before decimal point, carefully evaluate significant figures. mg/L

Explanation / Answer

(a) Write the reaction that occurs when H2SO4 is added to KIO3 + KI. (Use the lowest possible coefficients. Omit spectator ions and states-of-matter.)

5KI + KIO3 + 3H2SO4 3I2 + 3K2SO4 + 3H2O

(b) Write a balanced reaction between I3 and sulfite. (Use the lowest possible coefficients. Omit spectator ions and states-of-matter.)

SO3^2- + I3^- ---> SO4^2- + 3I^-

(c) Find the concentration of sulfite in the wine. Express your answer in mol/L and in mg SO32 per liter.
_____mol/L  HINTS: Don't use scientific notation, do put zero before decimal point, carefully evaluate significant figures.
_____mg/L

First calculate the mmoles of I3^- as follows:

[(0.8043 g /214 g/ mol ) /20] *(3 mmol I3^- /1 mmol IO3^-)*1000

= 0.564 mmol I3^-

Here 1/20 = 20/100 for dilution

Now calculate the excess I3^-= 12.86 *0.0481/ * 1 mol I3^-/ 2 moleS2O3= 0.310mole I3^-

Then we calculate the moles of I3^- which are reacted with SO3^2-= mmSO3^2- = 0.564 -0.310= 0.254 mmolSO3^2-

Molarity = 0.254 mmolSO3^2- /50 ml

= 2.54*10^-4 Mole /0.050 L

= 5.080*10^-3 M / mole /L

Now calculate the concentration into mg /L

Amount of SO3^2- in g = number of moles * molar mass

= 5.080*10^-3 mole /L * 80.0632 g/ mole

= 0.4067 g /L

= 406.7 mg/L

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