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Simora [160]
1 year ago
8

Write the chemical equation for fot the reaction between mgcl2 and the soap you preapred

Chemistry
1 answer:
In-s [12.5K]1 year ago
5 0
Soap is the sodium or potassium salt of long chain of fatty acid. Fatty acids when treated with NaOH or KOH forms Soap. This process is called as Saponification. Examples of Soap are as follow,

                                     1.  Sodium Stearate C₁₇H₃₅COONa
                                   
                                     2.  Potassium Oleate C₁₇H₃₃COOK

Reaction of Soap with MgCl₂;

When Soap is treated with MgCl₂ or CaCl₂ it forms insoluble precipitate called S.C.U.M. The reactions with MgCl₂ are as follow,

                2C₁₇H₃₅COONa + MgCl₂  -------->  2C₁₇H₃₅COOMg  + 2 NaCl

                2C₁₇H₃₃COOK   + MgCl₂  -------->  2C₁₇H₃₅COOMg  + 2 KCl

These reaction are often found in hard water. And this reaction decreases the effectiveness of soap.
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Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t
MariettaO [177]

Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

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