Review Test submissic eEEReview Test submissic ·Take Test: Homework As Renew Tes

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Review Test submissic eEEReview Test submissic ·Take Test: Homework As Renew Test submission Your question was ans ex . > C manual, resources Course Calendar Discussions Groups Tools f secure https:/ toro csudh.edu/webapps/assessment/take/takejsp?course assessment id-191547 1&course; id= 207422 1&content; id= 4447998 1 &question; num 5. Question Completion Status: Moving to another question will save this response Question 5 of 6 uestion 5 0.5 points Saved Back Titration similar to 16-14 100.00 mL solution). The winetreagent To 40.5 mL of wine were added 5 mL of concentrated Kl solution and 4.359 mL of reagent solution containing ( 0.898 KIO3) mixture was acidified with 10 mL of 1 M HCI to create a known number of moles of l2 by the reaction IO3. + excessl. + someH+-> 312 Some of this 12 immediately reacted with the sulfite in the wine sample by the reaction SO32- + 12 --> products we don't care about The left over l2 was titrated to a starch end point with 11.12 mL of 0.04299 M thiosulfate solution, by the reaction 12 + 2S2032--> products we don't care about What is the concentration of sulfite in the wine, expressed in ppm S032 wine? (mg S032 / Liter) > Mloving to another question will save this response. Question 5 of 6 (x ENG 10:14 PM 3/13/2018 Type here to search ^

Explanation / Answer

Moles of IO3- added = (0.898 / 100) x 4.359 = 0.03914

Moles of H+ added = 10 / 1000 = 0.01 (Limiting Reagent)

Moles of I2 produced = 3 x 0.01 = 0.03

Moles of Thiosulfate reacted = 0.04299 x (11.12/1000) = 0.000478

Moles of I2 reacted = 0.000478 / 2 = 0.000239

Moles of I2 reacted with SO32- = Moles of SO32- = 0.03 - 0.000239 = 0.029761

Concentration of sulfite in the wine = 0.029761 / 0.059859 = 0.49718 M

Concentration in ppm = 0.49718 x 80.0632 x 1000 = 39805 ppm

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