Question

Balance the following chemical equations:
I. $ Mg{{(OH)}_{2}}+HCl\to MgC{{l}_{2}}+{{H}_{2}}O$
II. $ {{N}_{2}}+{{H}_{2}}\to N{{H}_{3}}$
III. $ {{P}_{4}}+{{O}_{2}}\to {{P}_{2}}{{O}_{5}}$

Answer 1

Hint: In balancing, you have to make sure that the number of atoms of each element in the reactant side is equal to the number of atoms of each element in the product side.

Complete answer:
In order to answer our question, we need to learn about balancing and why it is used. Now, we know that mass can neither be created nor destroyed in a chemical reaction. Moreover, given a set of reactants, no new species of element can be formed in the products. For example, hydrogen and oxygen combine to form water, and water has both hydrogen and oxygen present in it. No new species such as nitrogen or argon has ended in the product.
The main rule in balancing is to keep the number of atoms of each element in the reactant side and the product side the same. Now, let us see the equations:
(i)In this equation, we have a problem with the hydrogen and oxygen atoms. So, we can add 2HCl in the reactant side and $ 2{{H}_{2}}O$ in the product side and doing this will make the number of atoms equal and the reaction will look like this:
\[Mg{{(OH)}_{2}}+2HCl\to MgC{{l}_{2}}+2{{H}_{2}}O\]
(II)In this equation, we see that there is a problem with the number of hydrogen. So, we can add $ 3{{H}_{2}}$ in the reactant side and $ 2N{{H}_{3}}$ in the product side to compensate and balance the equation. So the final equation looks like this:
\[{{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}\]
(III)In this equation the number of phosphorus atoms are not equal. In order to balance the phosphorus atoms we add $ 2{{P}_{2}}{{O}_{5}}$ in the product side, and to compensate for the oxygen, we add $ 5{{O}_{2}}$ in the reactant side. So, we obtain the final balanced equation as:
\[{{P}_{4}}+5{{O}_{2}}\to 2{{P}_{2}}{{O}_{5}}\]
Hence, we obtain the balanced chemical reactions for the given set of equations.

Note: It is to be noted that if the balanced chemical equations are multiplied or divided by a constant throughout all the elements, then also the equation remains balanced.