71.0k views
5 votes
What volume of 0.00143 M K2Cr2O7 is required to titrate 1.9 g of CH3OH dissolved in 50 ml of solution

User Vao Tsun
by
6.6k points

1 Answer

4 votes

Answer:

Step-by-step 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

User Sunil Kumar Jha
by
5.5k points