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
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Chapter4: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 47E
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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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