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A sink drain can become clogged with solid fat such as glyceryl tristearate (tristearin)

a. How would adding lye NaOHto the sink drain remove the blockage?

b. Write a balanced chemical equation for the reaction that occurs.

c. How many milliliters of a 0.500MNaOHsolution are needed to completely react with 10.0gof glyceryl tristearate (tristearin)?

Short Answer

Expert verified

(Part a) The sink reduces the lecithin tristearate since detergent is water based.

(Part b)

(Part c) To completely react with 10.0gof glyceryl tristearate (tristearin) is0.0672Lor67.2ml

Step by step solution

01

(Part a) Step 1: Introduction.

Lipids are found naturally components this included fats, waxes, and sterols. triglycerides, for illustration Cholesterol are fats that have included glycerol bonded to three fatty acids, such like fats and oils.

02

(Part a) Step 2: Given Information. 

Substance and glycerol salts are founded on the basis of all this procedure.

03

(Part a) Step 3: Explanation. 

(a). Heat the butter with only a good base for sodium hydroxide (NaOH) or potassium hydroxide KOHto saponify it.

Marinating with provides a smoother hand soap, however culturing and generates a hard cleanse.

Bringing lye NaOHto the sink reduces the lecithin tristearate since detergent is water based.

04

(Part b) Step 4: Given Information. 

Stearic acid is a linolenic acid acid with 18carbons and no deprotonation.

05

(Part b) Step 5: Explanation.

(b) The trying to follow is the esterification mechanism of glyceryl material creates inside this condition of KOH:

06

(Part c) Step 6: Given Information.

Three molecules of NaOHcombine for one molecule of glyceryl tristearate .

07

(Part c) Step 7: Explanation.

(c) Glyceryl tristearate has to have a detection limit of 891.48g/mol.

The molecules of glyceryl typically are

= Weight/molecular weight

=10g/891.48

=0.0112M.

. The molar mass of 10mgof glyceryl material creates is

Three moles NaOHreact through one mole of glyceryl tristearate. As an outcome, 0.0112Mlecithin tristearate.

=0.0336/0.5

=0.0672Lor67.2ml

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