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Simora [160]
3 years ago
8

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

Chemistry
1 answer:
In-s [12.5K]3 years 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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3 0
2 years ago
A compound of a transition metal and iodine is 56.7% metal by mass.How many grams of the metal can be obtained from 630 g of thi
erma4kov [3.2K]

Answer:

357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine

Explanation:

In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal  present in 630 g of the compound is approximately 357 g

4 0
3 years ago
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
3 years ago
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