Question

Write the balanced chemical equations for the following reactions.
A.$Calcium\,hydroxide+Carbon\,dioxide\to Calcium\,carbonate+Water$
B.$Zinc+Silver\,Nitrate\to Zinc\,Nitrate+Silver$
C.$Alu\min ium+Copper\,chloride\to Alu\min ium\,chloride+Copper$
D.$Barium\,chloride+Potassium\,sulphate\to Barium\,sulphate+Potassium\,chloride$

Answer 1

Hint In order to balance the chemical equations, the number of atoms of each specific element in the reactant side should be equal to the number of elements of that element in the product side too.

Complete step by step solution:
In order to answer our question, we need to learn about the laws of chemical combination. Now, we know that in any chemical reaction, energy can neither be created nor destroyed. The same applies to mass. Mass remains constant, i.e total mass of the reactants is equal to the mass of the products. Also, no new element can suddenly come out in the reaction. For example in the reaction of ${{H}_{2}}$ and ${{O}_{2}}$ to form ${{H}_{2}}O$, the only elements present are hydrogen and water. No new element like nitrogen has entered the reaction. Moreover, the number of atoms of each element should be the same in the reactant and the product side. Now, let us balance each of the above reactions:
(A). In this reaction, calcium hydroxide is reacting with carbon dioxide to produce calcium carbonate along with water. The reaction can be represented chemically as:
$Ca{{(OH)}_{2}}+C{{O}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O$
We can see that the reaction is already balanced as a number of atoms of each element in the reactant and the product side remain the same.
(B)In this reaction, zinc reacts with silver nitrate to form zinc nitrate and silver. After writing the reaction, we see there is an imbalance of atoms of Ag, so we multiply $AgN{{O}_{3}}$ and Ag by a factor of 2 and the final reaction is:
$Zn+2AgN{{O}_{3}}\to Zn{{(N{{O}_{3}})}_{2}}+2Ag$
(C)In this reaction, aluminium reacts with copper chloride, and none of the atoms of the elements is balanced. By trial and error method, we balance the atoms and the final reaction looks like:
$2Al+3CuC{{l}_{2}}\to 2AlC{{l}_{3}}+3Cu$
(D)In this reaction, we can observe that the potassium and chlorine atoms are not balanced. However, if we multiply $KCl$ by a factor of 2, then the overall reaction gets balanced and it looks like:
$BaC{{l}_{2}}+{{K}_{2}}S{{O}_{4}}\to BaS{{O}_{4}}+2KCl$
Hence, we have obtained all the balanced chemical reactions.

NOTE: If we multiply the balanced chemical reactions by a constant throughout, even then the reaction remains balanced.