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0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was?
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0.01102 M HOl. Calculate the sulfur concentration in the sample in par...
Calculation of Sulfur Concentration in the Sample
- Given: 0.01102 M HOl
- The balanced chemical equation for the reaction of sulfur with HOl is:
S + 2HOl → H2SO4
- From the equation, we can see that 1 mole of S reacts with 2 moles of HOl to form 1 mole of H2SO4.
- Therefore, the concentration of sulfur in the sample is:
0.01102 M HOl x (1 mole S/2 moles HOl) = 0.00551 M S
- To convert this to parts per million (ppm), we can use the following formula:
ppm = (mass of component/mass of sample) x 1,000,000
- Assuming a sample size of 100 mL and a density of water of 1 g/mL, the mass of the sample is:
100 mL x 1 g/mL = 100 g
- The molar mass of sulfur is 32.06 g/mol, so the mass of sulfur in the sample is:
0.00551 M S x 32.06 g/mol x 0.1 L = 0.0176 g S
- Therefore, the sulfur concentration in the sample is:
(0.0176 g S/100 g sample) x 1,000,000 = 176 ppm
Calculation of Iron Concentration in the Sample
- Given: 100.0 mL of spring water, 25.00 mL of 0.002517 M K2CrO4, 8.53 mL of 0.00949 M Fe2 solution, and the balanced chemical equation:
6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
- The K2CrO4 reacts with Fe2+ to form Fe3+, which is then back-titrated with Fe2+.
- From the balanced equation, we can see that 6 moles of Fe2+ react with 1 mole of Cr2O72- to form 6 moles of Fe3+.
- Therefore, the moles of Fe2+ in the sample are:
0.002517 M K2CrO4 x 0.02500 L = 6.2925 x 10^-5 moles Cr2O72-
6.2925 x 10^-5 moles Cr2O72- x (6 moles Fe2+/1 mole Cr2O72-) = 3.7755 x 10^-4 moles Fe2+
- The moles of Fe2+ in the back-titration are:
0.00949 M Fe2+ x 0.00853 L = 8.0887 x 10^-5 moles Fe2+
- Therefore, the moles of Fe2+ in the sample are:
3.7755 x 10^-4 moles Fe2+ - 8.0887 x 10^-5 moles Fe2+ = 2.9667 x 10^-4 moles Fe2+
- To convert this to parts per million (ppm), we can use the same formula as before:
ppm = (mass of component/mass of sample) x 1,000,
- Given: 0.01102 M HOl
- The balanced chemical equation for the reaction of sulfur with HOl is:
S + 2HOl → H2SO4
- From the equation, we can see that 1 mole of S reacts with 2 moles of HOl to form 1 mole of H2SO4.
- Therefore, the concentration of sulfur in the sample is:
0.01102 M HOl x (1 mole S/2 moles HOl) = 0.00551 M S
- To convert this to parts per million (ppm), we can use the following formula:
ppm = (mass of component/mass of sample) x 1,000,000
- Assuming a sample size of 100 mL and a density of water of 1 g/mL, the mass of the sample is:
100 mL x 1 g/mL = 100 g
- The molar mass of sulfur is 32.06 g/mol, so the mass of sulfur in the sample is:
0.00551 M S x 32.06 g/mol x 0.1 L = 0.0176 g S
- Therefore, the sulfur concentration in the sample is:
(0.0176 g S/100 g sample) x 1,000,000 = 176 ppm
Calculation of Iron Concentration in the Sample
- Given: 100.0 mL of spring water, 25.00 mL of 0.002517 M K2CrO4, 8.53 mL of 0.00949 M Fe2 solution, and the balanced chemical equation:
6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
- The K2CrO4 reacts with Fe2+ to form Fe3+, which is then back-titrated with Fe2+.
- From the balanced equation, we can see that 6 moles of Fe2+ react with 1 mole of Cr2O72- to form 6 moles of Fe3+.
- Therefore, the moles of Fe2+ in the sample are:
0.002517 M K2CrO4 x 0.02500 L = 6.2925 x 10^-5 moles Cr2O72-
6.2925 x 10^-5 moles Cr2O72- x (6 moles Fe2+/1 mole Cr2O72-) = 3.7755 x 10^-4 moles Fe2+
- The moles of Fe2+ in the back-titration are:
0.00949 M Fe2+ x 0.00853 L = 8.0887 x 10^-5 moles Fe2+
- Therefore, the moles of Fe2+ in the sample are:
3.7755 x 10^-4 moles Fe2+ - 8.0887 x 10^-5 moles Fe2+ = 2.9667 x 10^-4 moles Fe2+
- To convert this to parts per million (ppm), we can use the same formula as before:
ppm = (mass of component/mass of sample) x 1,000,
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0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was?
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0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was?.
0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was?.
Solutions for 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? in English & in Hindi are available as part of our courses for Chemical Engineering.
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Here you can find the meaning of 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? defined & explained in the simplest way possible. Besides giving the explanation of
0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was?, a detailed solution for 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? has been provided alongside types of 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? theory, EduRev gives you an
ample number of questions to practice 0.01102 M HOl. Calculate the sulfur concentration in the sample in parts per million. 21. A 100.0-mL sample of spring water was treated to convert any iron present to Fe? . Addition of 25.00 mL of 0.002517 M K2CrO, resulted in the reaction 6Fe? C1207?- 14H - › 6FeSt 2C13 7H10 The excess K2C1zO, was back-titrated with 8.53 mL of 0.00949 M Fe2 solution. Calculate the concentration of iron in the sample in parts per million. 22. The arsenic in a 1.203-g sample of a pesticide was converted to HzAsO4 by suitable treatment. The acid was then neutralized, and 40.00 mL of 0.05871 M AgNO, was added to precipitate the arsenic quantitatively as AgzAsO4. The excess Ag* in the filtrate and in the washings from the precipitate was? tests, examples and also practice Chemical Engineering tests.
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