Chemistry Examples

$K_{3} P O_{4} + B a {(N O_{3})}_{2} \to K N O_{3} + B a_{3} {(P O_{4})}_{2}$

Step 1

The law of conservation of mass states that matter cannot be created nor destroyed. There are no more or less atoms at the end of a chemical reaction than there were at the beginning. To balance a chemical equation, every element must have the same number of atoms on each side of the equation.

To balance a chemical equation:

1. Every element must have the same number of atoms on each side of the equation (law of conservation of mass).

2. Balancing the equation can be done only by adjusting the coefficients.

Step 2

$K$ is not a balanced element. To balance $K$ , multiply $K N O_{3}$ from the right side by $3$ .

$K_{3} P O_{4} + B a {(N O_{3})}_{2} \to 3 K N O_{3} + B a_{3} {(P O_{4})}_{2}$

Step 3

$P$ is not a balanced element. To balance $P$ , multiply $K_{3} P O_{4}$ from the left side by $2$ .

$2 K_{3} P O_{4} + B a {(N O_{3})}_{2} \to 3 K N O_{3} + B a_{3} {(P O_{4})}_{2}$

Step 4

$O$ is not a balanced element. To balance $O$ , multiply $K_{3} P O_{4}$ from the left side by $2$ .

$2 K_{3} P O_{4} + B a {(N O_{3})}_{2} \to 6 K N O_{3} + B a_{3} {(P O_{4})}_{2}$

Step 5

$B a$ is not a balanced element. To balance $B a$ , multiply $B a {(N O_{3})}_{2}$ from the left side by $3$ .

$2 K_{3} P O_{4} + 3 B a {(N O_{3})}_{2} \to 6 K N O_{3} + B a_{3} {(P O_{4})}_{2}$

Step 6

$N$ is a balanced element because there is the same number of atoms of $N$ in each side of the equation.

$N$ is a balanced element

Step 7

$O$ is a balanced element because there is the same number of atoms of $O$ in each side of the equation.

$O$ is a balanced element

Step 8

All elements have the same number of atoms in each side of the equation, which means that the law of conservation of mass is satisfied and the chemical equation is balanced.

$2 K_{3} P O_{4} + 3 B a {(N O_{3})}_{2} \to 6 K N O_{3} + B a_{3} {(P O_{4})}_{2}$