Question

Find the product formed $2NaOH + MgS{O_4} \to $
A) $MgO + N{a_2}S{O_4}$
B) $Mg{\left( {OH} \right)_2} + N{a_2}S{O_4}$
C) $Mg{\left( {OH} \right)_2}$
D) $MgO + N{a_2}O + {H_2}O$

Answer 1

Hint: Now we observe the double decomposition reaction.
The reaction where the two reactants trade their positive and negative particles and structure another compound is known as a twofold decay reaction and furthermore called double decomposition reaction. The overall condition of the reaction is,
$AB + CD \to AD + CB$

Complete step by step answer:
Let us see a case of a double decomposition reaction.
On warming, ferrous sulfate precious stones lose water and it structures anhydrous ferrous sulfate so the shade of the gem changes from light green to white, and on additional warming anhydrous ferrous sulfate disintegrates to give ferric oxide, sulfur dioxide, and sulfur trioxide
$2FeS{O_4}\xrightarrow{\Delta }F{e_2}{O_3} + S{O_2} + S{O_3}$
Sodium hydroxide and magnesium sulfate will respond to give an insoluble product slurry of \[Mg{\left( {OH} \right)_2}\left( s \right)\]. Sodium hydroxide is dissolvable in water, and dissimilar to the next group 2 sulfates, \[MgS{O_4}\] is solvent in water. At the point when blended you will have\[N{a^ + }\], \[O{H^ - }\], \[M{g^{2 + }}\] and \[S{O_4}^{2 - }\]. \[M{g^{2 + }}\] will respond with \[O{H^ - }\] to make \[Mg{\left( {OH} \right)_2}\left( s \right)\] which is insoluble. It's the stuff of Milk of Magnesia. Sulfate particles don't hydrolyze to any quantifiable degree.
The reaction is,
$2NaOH + MgS{O_4} \to Mg\left( {O{H_2}} \right)$$+N{a_2}S{O_4}$
The product of this above is $Mg{\left( {OH} \right)_2}$,$N{a_2}S{O_4}$

So, the correct answer is Option B.

Additional Note:
We have to remember that the neutralization and precipitation responses are some more instances of twofold deterioration responses.
Neutralization:
We must know that the neutralization might be a twofold deterioration response that happens between the acids and bases. The response among vinegar and bicarbonate of pop is a case of balance.
Precipitation response:
We have to know that a precipitation response might be a twofold decay that happens between two fluid ionic mixes which structure a substitution insoluble ionic compound. The response between lead (II) nitrate and iodide is a case of a precipitation response.

Note: Now we discuss about how the chemical displacement and decomposition reaction are differ from each other,
Chemical displacement:
In the synthetic response on the off chance that one metal dislodges to another compound structure another item called substance displacement response.
Example: \[Zn + 2HCl \to ZnC{l_2} + {H_2}\]
Chemical decomposition:
The compound reaction in which reactants decays into at least two items called chemical decomposition.
Example: $CaC{O_3}\xrightarrow{\Delta }CaO + C{O_2}$.