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1. One menthol used commercially to peel potatoes is to soak them in a solutionof NaOH for a short time, remove them from the NaOH and spray off the peel. The NaOH is analyzed periodically. In one such analysis, 40.72 mL of 0.400 M H2SO4 is required to react completely with a 20.0 mL sample of the NaOH solution. The reaction involved is: 2 KOH + H2SO4 -> K2SO4 + 2 H2O a. How many moles of H2SO4 were used in the analysis (0.040722) 20.4 mol ) b. Based on the equation, How is NaOH stoichiometrically equivalent to H2SO4 2:1 2 NaOH + H₂Soy Na₂ Soy + 2 H₂O → c. How many moles of NaOH is present in the solution d. What is the molarity of the NaOH 0.016288 or 0.02 moles

1. One menthol used commercially to peel potatoes is to soak them in a solutionof NaOH for a short time, remove them from the NaOH and spray off the peel. The NaOH is analyzed periodically. In one such analysis, 40.72 mL of 0.400 M H2SO4 is required to react completely with a 20.0 mL sample of the NaOH solution. The reaction involved is: 2 KOH + H2SO4 -> K2SO4 + 2 H2O a. How many moles of H2SO4 were used in the analysis (0.040722) 20.4 mol ) b. Based on the equation, How is NaOH stoichiometrically equivalent to H2SO4 2:1 2 NaOH + H₂Soy Na₂ Soy + 2 H₂O → c. How many moles of NaOH is present in the solution d. What is the molarity of the NaOH 0.016288 or 0.02 moles

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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1. One menthol used commercially to peel potatoes is to soak them in a solutionof NaOH for a short time, remove them from the NaOH and spray off the peel. The NaOH is analyzed periodically. In one such analysis, 40.72 mL of 0.400 M H2SO4 is required to react completely with a 20.0 mL sample of the NaOH solution. The reaction involved is: 2 KOH + H2SO4 -> K2SO4 + 2 H2O a. How many moles of H2SO4 were used in the analysis mol (0.04072 2) (0.4m02) = 0.016288 or 0.02 moles b. Based on the equation, How is NaOH stoichiometrically equivalent to H2SO4 2:1 2 NaOH + H₂SO4 → Na₂ Soy + 2 H ₂ O c. How many moles of NaOH is present in the solution d. What is the molarity of the NaOH 2. A solution of 0.255 N KOH is needed to analyze acidic mine water. K=39, O=16, H=1 a. Calculate the mass of KOH needed to prepare 500mL of the solution mol kot 20.255 mol ) (0.5x) = 0.1275 mol Kot 9 кон = 0. 1275 X mol 56 g/mol = 7.14 g KOH I mol x=39 0 = 16 H = 1 = 56 g/mol 3. 0.24 N HCI is needed to analyze a sample of soda ash a. Calculate the volume of conc. HCI to prepare 750 mL of the solution, Conc. HCI = 12 M C₁V₁ =C₂V₂ (0.24 M) (750mm) (1215) (V₂) 12, 19 12149 √₂-15 ml of HC1
Transcribed Image Text:1. One menthol used commercially to peel potatoes is to soak them in a solutionof NaOH for a short time, remove them from the NaOH and spray off the peel. The NaOH is analyzed periodically. In one such analysis, 40.72 mL of 0.400 M H2SO4 is required to react completely with a 20.0 mL sample of the NaOH solution. The reaction involved is: 2 KOH + H2SO4 -> K2SO4 + 2 H2O a. How many moles of H2SO4 were used in the analysis mol (0.04072 2) (0.4m02) = 0.016288 or 0.02 moles b. Based on the equation, How is NaOH stoichiometrically equivalent to H2SO4 2:1 2 NaOH + H₂SO4 → Na₂ Soy + 2 H ₂ O c. How many moles of NaOH is present in the solution d. What is the molarity of the NaOH 2. A solution of 0.255 N KOH is needed to analyze acidic mine water. K=39, O=16, H=1 a. Calculate the mass of KOH needed to prepare 500mL of the solution mol kot 20.255 mol ) (0.5x) = 0.1275 mol Kot 9 кон = 0. 1275 X mol 56 g/mol = 7.14 g KOH I mol x=39 0 = 16 H = 1 = 56 g/mol 3. 0.24 N HCI is needed to analyze a sample of soda ash a. Calculate the volume of conc. HCI to prepare 750 mL of the solution, Conc. HCI = 12 M C₁V₁ =C₂V₂ (0.24 M) (750mm) (1215) (V₂) 12, 19 12149 √₂-15 ml of HC1
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