Question

In a volumetric analysis experiment, an acidic aqueous solution of methanol (CH3OH) is titrated with a solution of potassium dichromate (K2Cr2O7) according to thefollowing balanced chemical equation:
2K2Cr2O7(aq) + 8H2SO4(aq) + 3CH3OH(aq) ===> 2Cr2(SO4)3(aq) + 11H2O(l) + 3HCOOH(aq) + 2K2SO4(aq)
It required 24.14 mL of 0.0357 M K2Cr2O7 to reach the endpoint. What mass of CH3OH was present initially? Please Help help :)) Thanks in advance.

Answer 1

.02414L * .0357M=mol of K2Cr2O7 *(this is the mole ratio)=mol CH3OH * (g/mol) of CH3OH

Answer 2

Moles of K2Cr2O7 = volume x concentration of K2Cr2O7

= 24.14/1000 x 0.0357 = 0.0008618 mol

Moles of CH3OH = 3/2 x moles of K2Cr2O7 (from reaction equation)

= 3/2 x 0.0008618 = 0.001293 mol

Mass of CH3OH = moles x molar mass of CH3OH

= 0.001293 x 32.04

= 0.0414 g

Answer 3