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

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

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

the answer is 75.0%.