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

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.