Question

Complete the below given reaction-
$Ca{\left( {HC{O_3}} \right)_2} + 2HCl \to $
A.$CaC{l_2} + 2{H_2}O + 2C{O_2}$
B.$2CaC{l_2} + 2{H_2}O + 2C{O_2}$
C.$CaC{l_2} + 4{H_2}O + 2C{O_2}$
D.$CaC{l_2} + 2{H_2}O + 4C{O_2}$

Answer 1

Hint: We know that $Ca{\left( {HC{O_3}} \right)_2}$ is calcium carbonate and $HCl$ is hydrochloric acid. We have to know that $Ca{\left( {HC{O_3}} \right)_2}$ is metal carbonate and $HCl$ is acid. Generally, we can say that metal carbonates are bases. We have to know that when acid and bases react with each other, products formed would be salt and water.

Complete answer:
We saw that when an acid and carbonate reacts with one other, the products formed will be water, salt, and carbon dioxide. We can write the general equation as,
$Acid + Metalcarbonate \to Salt + Water + Carbondioxide$
In the given equation, the metal carbonate is calcium carbonate. We can write the chemical formula of the metal carbonate as $Ca{\left( {HC{O_3}} \right)_2}$ and the acid is hydrochloric acid, the formula of acid is $HCl$.
So, when calcium carbonate reacts with hydrochloric acid, the products formed would be calcium chloride, water, and carbon dioxide. We can write the chemical equation as,
$Ca{\left( {HC{O_3}} \right)_2} + HCl \to CaC{l_2} + {H_2}O + C{O_2}$
We can see that the above equation is unbalanced. We need two moles of hydrochloric acid in the reactant side, two moles of water, and two moles of carbon dioxide in the product side to make the reaction a balanced one.
Let us now write the balanced equation as,
$Ca{\left( {HC{O_3}} \right)_2} + HCl \to CaC{l_2} + 2{H_2}O + 2C{O_2}$

Option (A) is correct.

Note:
We have to know that during this reaction, bubbling is caused by carbon dioxide and it could be identified by passing gas by lime water that would turn it cloudy. We have to know that the reaction between metal carbonate and acid is exothermic which means energy of heat is acidic. We have to know that these kinds of reactions could be utilized to identify if the unknown solutions are acidic.