10.0 mL of 0.00500 M I2, and then back-titrating the excess I2 with 0.00200 M Na2S2O3. If 2.6 mL Na2S2O3 is requiredfor the titration, how many milligrams of sulfur are contained in the sampleReactions:H2S + I2 → S + 2I? + 2H+I2 + 2 S2O32? → 2I? + S4O6
H2S + I2 → S + 2 I{-} + 2 H{+} I2 + 2 S2O3{2-} → 2 I{-} + S4O6 (I'm going to ignore the fact that the charges don't balance in your second equation. It's a lot of trouble to sort it out when the usual equation for that is given as: I3{-} + 2 S2O3{2-} → 3 I{-} + S4O6{2-} where I3{-} is the result of adding I2 to I{-}.) (0.00200 mol Na2S2O3/L) × (0.0026 L Na2S2O3) × (1 mol I2 / 2 mol Na2S2O3) 2.6 × 10^-6 mol I2 reacted with Na2S2O3 (0.0100 L) × (0.00500 mol I2/L) 5.0 × 10^-5 mol I2 total (5.0 × 10^-5 mol I2 total) - (2.6 × 10^-6 mol I2 reacted with Na2S2O3) 4.74 × 10^-5 mol I2 reacted with H2S (4.74 × 10^-5 mol I2) × (1 mol H2S / 1 mol I2) x (1 mol S / 1 mol H2S) × (32.0655 g S/mol) 0.00152 g 1.52 mg S
0.00200 M Na2S2O3. X 2.6 mL Na2S2O3 A mmole Na2S2O3. . this is the source of the (S2O3)- ion in the 2nd reaction A mmole (S2O3)-. X (1 mmol I2 / 2 mmole (S2O3)- ) (A/2) mmole I2 this is the excess that was back titrated 10.0 mL X 0.00500 M I2 0.0500 mmole I2 available for reaction [0.0500 - (A/2)] mmole I2 amount used in the first reaction . mole ration in 1st reaction is 1 mole S / 1 mole I2 so the amounts used are equal . [0.0500 - (A/2)] mmole I2 will react with [0.0500 - (A/2)] mmole S . Starting at the top of my explanation . 1. calculate A . it should be 0.0052 mmole . 2. find [0.0500 - (A/2)] . it should be 0.00474 mmol last use atomic mass of S to find the final answer . 0.00474mmol X 32.06 mg/mmole I leave this last calculation for you
H2S + I2 → S + 2 I{-} + 2 H{+} I2 + 2 S2O3{2-} → 2 I{-} + S4O6 (I'm going to ignore the fact that the charges don't balance in your second equation. It's a lot of trouble to sort it out when the usual equation for that is given as: I3{-} + 2 S2O3{2-} → 3 I{-} + S4O6{2-} where I3{-} is the result of adding I2 to I{-}.) (0.00200 mol Na2S2O3/L) × (0.0026 L Na2S2O3) × (1 mol I2 / 2 mol Na2S2O3) 2.6 × 10^-6 mol I2 reacted with Na2S2O3 (0.0100 L) × (0.00500 mol I2/L) 5.0 × 10^-5 mol I2 total (5.0 × 10^-5 mol I2 total) - (2.6 × 10^-6 mol I2 reacted with Na2S2O3) 4.74 × 10^-5 mol I2 reacted with H2S (4.74 × 10^-5 mol I2) × (1 mol H2S / 1 mol I2) x (1 mol S / 1 mol H2S) × (32.0655 g S/mol) 0.00152 g 1.52 mg S
Can you afford to go to a dermatologist?
Can you afford to go to a dermatologist?
0.00200 M Na2S2O3. X 2.6 mL Na2S2O3 A mmole Na2S2O3. . this is the source of the (S2O3)- ion in the 2nd reaction A mmole (S2O3)-. X (1 mmol I2 / 2 mmole (S2O3)- ) (A/2) mmole I2 this is the excess that was back titrated 10.0 mL X 0.00500 M I2 0.0500 mmole I2 available for reaction [0.0500 - (A/2)] mmole I2 amount used in the first reaction . mole ration in 1st reaction is 1 mole S / 1 mole I2 so the amounts used are equal . [0.0500 - (A/2)] mmole I2 will react with [0.0500 - (A/2)] mmole S . Starting at the top of my explanation . 1. calculate A . it should be 0.0052 mmole . 2. find [0.0500 - (A/2)] . it should be 0.00474 mmol last use atomic mass of S to find the final answer . 0.00474mmol X 32.06 mg/mmole I leave this last calculation for you
