Question

Balance the following chemical equations:
a. $HN{O_3} + Ca{(OH)_2} \to Ca{(N{O_3})_2} + {H_2}O$
b. $NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O$
c. $NaCl + AgN{O_3} \to AgCl + NaN{O_3}$
d. $BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + HCl$

Answer 1

Hint: Use the inspection method to find the difference of number of atoms in the reactants and products side. By conservation of matter, the number of atoms in a reaction must be constant and same on reactants and products side. Based on the difference, multiply the molecules by appropriate coefficients. Do this till all atoms are balanced out.

Complete step by step solution:
We will consider the first chemical equations:
a. $HN{O_3} + Ca{(OH)_2} \to Ca{(N{O_3})_2} + {H_2}O$
Now we will count the number of atoms from each molecule on the reactants side and compare them with the corresponding number on the products side. It is better to follow a step by step method of visual inspection and balancing atoms one by one. There is no standard theoretical procedure for doing this exercise. Now we see the $Ca$ atoms are balanced on both sides, but on the reactants side, we see that there is one $N$ atom whereas there are two of them on the products side. So we will multiply the first reactant by a coefficient 2 then we get,
$2HN{O_3} + Ca{(OH)_2} \to Ca{(N{O_3})_2} + {H_2}O$
Let us calculate the number of $O$ atoms on both sides. Reactants = (2 $\times$ 3)+(2 $\times$ 1) = 6+2 = 8. In the products side = (2 $\times$ 3)+1 = 6+1 = 7. So there is one $O$ on the product side less than the reactants. We will multiply the second reactant with a coefficient, 2. Thus we will get, as follows:
$2HN{O_3} + Ca{(OH)_2} \to Ca(N{O_3}) + 2{H_2}O$
Now we see that all the atoms on both sides of the reaction are balanced. This is the final balanced chemical equation.

Now consider the second chemical equation:
b. $NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O$
After a visual inspection we see that there is one atom of $Na$ on the reactants side and two atoms on the products side. So we will multiply the first product by 2 as a coefficient. Then we get,
$2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O$
Now we calculate the number of $O$ atoms on both sides. Reactants = (2 $\times$ 1)+4 = 2+4 = 6. In the products side = 4+1=5. Thus there is one less $O$ on the products side than compared to the reactants side. We will multiply second product with 2 as a coefficient, then we get,
$2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O$
Above equation is a balanced chemical equation with equal numbers of each atom on both sides.

c. $NaCl + AgN{O_3} \to AgCl + NaN{O_3}$
After a visual inspection we see that all the atoms are completely balanced with equal numbers on both sides of reaction. This equation with no modification is a balanced chemical equation.

d. $BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + HCl$
After a visual inspection we see that there are two atoms each of $Cl$ and $H$ on the reactants side but only one atom each on the products side. Fortunately we also have one molecule of $HCl$ consisting of both atoms. So we will multiply the second product with 2 as coefficient. This gives,
$BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl$
Above equation is a balanced chemical equation as all the atoms have the same number on both sides of the reaction.

Note: As mentioned in the solution above there is no standard procedure but only practice will help develop some tricks to notice during this exercise. One suggestion would be to start with the atoms with lowest number generally the metal atoms and then moving towards the lighter atoms like $H, O$ which are largely present in most chemical equations.