In a volumetric analysis experiment, an acidic aqueous solution of methanol (CH3OH) is titrated with a solution of potassium dichromate (K2Cr2O7) according to the following balanced chemical equation: 2K2Cr2O7 (aq) + 8H2SO4 (aq) + 3CH3OH (aq) → 2Cr2 (SO4) 3 (aq) + 11H2O (l) + 3 HCOOH (aq) + 2K2SO4 (aq) What volume of 0.00143 M K2Cr2O7 is required to titrate 1.90 g of CH3OH dissolved in 50.0 mL of solution

Question

Aniyah Bullock

Chemistry

21 November, 13:13

In a volumetric analysis experiment, an acidic aqueous solution of methanol (CH3OH) is titrated with a solution of potassium dichromate (K2Cr2O7) according to the following balanced chemical equation: 2K2Cr2O7 (aq) + 8H2SO4 (aq) + 3CH3OH (aq) → 2Cr2 (SO4) 3 (aq) + 11H2O (l) + 3 HCOOH (aq) + 2K2SO4 (aq) What volume of 0.00143 M K2Cr2O7 is required to titrate 1.90 g of CH3OH dissolved in 50.0 mL of solution

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April · Answer 1

27.64 liters

Explanation:

From the balanced equation, 2 moles of K2Cr2O7 requires 3 moles of CH3OH.

Mole of CH3OH = 1.9/32.04 = 0.0593 mole

Mole of K2Cr2O7 that will require 0.0593 mole of CH3OH:

2 x 0.0593/3 = 0.0395 mole

mole = molarity x volume

Volume of K2Cr2O7 needed = 0.0395/0.00143

= 27.64 Liter

Hence, 27.64 liters of 0.00143 M K2Cr2O7 will be required to titrate 1.90 g of CH3OH dissolved in 50.0 mL of solution