What volume of 0.00143 M K2Cr2O7 is required to titrate 1.9 g of CH3OH dissolved in 50 ml of solution

Explanation:

2K₂Cr₂O₇ + 8H₂SO₄ + 3CH₃OH=  2 K₂SO₄ + 2Cr₂( SO₄)₃ + 3 HCOOH + 11 H₂O .

1.9 g of CH₃OH = 1.9 / 32 moles of CH₃OH

moles of given CH₃OH = .059375 moles

3 moles of CH₃OH reacts with 2 moles of K₂Cr₂O₇  

.059375 moles of CH₃OH reacts with 2x .059375 / 3  moles of K₂Cr₂O₇  

Moles of K₂Cr₂O₇   required =  2x .059375 / 3   = .039583 moles

Let V be the volume in litre required

V x .00143 = .039583

V = 27.68 L

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