Chemistry Examples

$K_{4} (F e {(C N)}_{6}) + K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to F e_{2} {(S O_{4})}_{3} + C O_{2} + H N O_{3} + C r_{2} (S O_{4}) + K_{2} S O_{4} + H_{2} O$

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 $K_{2} S O_{4}$ from the right side by $3$ .

$K_{4} (F e {(C N)}_{6}) + K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to F e_{2} {(S O_{4})}_{3} + C O_{2} + H N O_{3} + C r_{2} (S O_{4}) + 3 K_{2} S O_{4} + H_{2} O$

Step 3

$F e$ is not a balanced element. To balance $F e$ , multiply $K_{4} (F e {(C N)}_{6})$ from the left side by $2$ .

$2 K_{4} (F e {(C N)}_{6}) + K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to F e_{2} {(S O_{4})}_{3} + C O_{2} + H N O_{3} + C r_{2} (S O_{4}) + 3 K_{2} S O_{4} + H_{2} O$

Step 4

$C$ is not a balanced element. To balance $C$ , multiply $C O_{2}$ from the right side by $12$ .

$2 K_{4} (F e {(C N)}_{6}) + K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to F e_{2} {(S O_{4})}_{3} + 12 C O_{2} + H N O_{3} + C r_{2} (S O_{4}) + 3 K_{2} S O_{4} + H_{2} O$

Step 5

$N$ is not a balanced element. To balance $N$ , multiply $H N O_{3}$ from the right side by $12$ .

$2 K_{4} (F e {(C N)}_{6}) + K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to F e_{2} {(S O_{4})}_{3} + 12 C O_{2} + 12 H N O_{3} + C r_{2} (S O_{4}) + 3 K_{2} S O_{4} + H_{2} O$

Step 6

$K$ is not a balanced element. To balance $K$ , multiply $K_{2} C r_{2} O_{7}$ from the left side by $3$ and $3 K_{2} S O_{4}$ from the right side by $5$ .

$2 K_{4} (F e {(C N)}_{6}) + 3 K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to F e_{2} {(S O_{4})}_{3} + 12 C O_{2} + 12 H N O_{3} + C r_{2} (S O_{4}) + 15 K_{2} S O_{4} + H_{2} O$

Step 7

$C r$ is not a balanced element. To balance $C r$ , multiply $C r_{2} (S O_{4})$ from the right side by $3$ .

$2 K_{4} (F e {(C N)}_{6}) + 3 K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to F e_{2} {(S O_{4})}_{3} + 12 C O_{2} + 12 H N O_{3} + 3 C r_{2} (S O_{4}) + 15 K_{2} S O_{4} + H_{2} O$

Step 8

$O$ is not a balanced element. To balance $O$ , multiply $K_{2} C r_{2} O_{7}$ from the left side by $61$ and $F e_{2} {(S O_{4})}_{3}, 12 C O_{2}, 12 H N O_{3}, 3 C r_{2} (S O_{4}), 15 K_{2} S O_{4}, H_{2} O$ from the right side by $5$ .

$2 K_{4} (F e {(C N)}_{6}) + 61 K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + 60 C O_{2} + 60 H N O_{3} + 15 C r_{2} (S O_{4}) + 75 K_{2} S O_{4} + 5 H_{2} O$

Step 9

$H$ is not a balanced element. To balance $H$ , multiply $H_{2} S O_{4}$ from the left side by $35$ .

$2 K_{4} (F e {(C N)}_{6}) + 61 K_{2} C r_{2} O_{7} + 35 H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + 60 C O_{2} + 60 H N O_{3} + 15 C r_{2} (S O_{4}) + 75 K_{2} S O_{4} + 5 H_{2} O$

Step 10

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

$2 K_{4} (F e {(C N)}_{6}) + 61 K_{2} C r_{2} O_{7} + 105 H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + 60 C O_{2} + 60 H N O_{3} + 15 C r_{2} (S O_{4}) + 75 K_{2} S O_{4} + 5 H_{2} O$

Step 11

$O$ is not a balanced element. To balance $O$ , multiply $H_{2} S O_{4}$ from the left side by $157$ and $5 F e_{2} {(S O_{4})}_{3}, 60 C O_{2}, 60 H N O_{3}, 15 C r_{2} (S O_{4}), 75 K_{2} S O_{4}, 5 H_{2} O$ from the right side by $127$ .

$2 K_{4} (F e {(C N)}_{6}) + 61 K_{2} C r_{2} O_{7} + 157 H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + 60 C O_{2} + 60 H N O_{3} + 61 C r_{2} (S O_{4}) + 81 K_{2} S O_{4} + 127 H_{2} O$

Step 12

$K$ is not a balanced element. To balance $K$ , multiply $2 K_{4} (F e {(C N)}_{6})$ from the left side by $10$ and from the right side by $65$ .

$10 K_{4} (F e {(C N)}_{6}) + 61 K_{2} C r_{2} O_{7} + 157 H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + 60 C O_{2} + 60 H N O_{3} + 61 C r_{2} (S O_{4}) + 81 K_{2} S O_{4} + 127 H_{2} O$

Step 13

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

$F e$ is a balanced element

Step 14

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

$C$ is a balanced element

Step 15

$N$ 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 16

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

$K$ is a balanced element

Step 17

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

$C r$ is a balanced element

Step 18

$O$ 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 19

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

$H$ is a balanced element

Step 20

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

$S$ is a balanced element

Step 21

$O$ 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 22

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.

$10 K_{4} (F e {(C N)}_{6}) + 61 K_{2} C r_{2} O_{7} + 157 H_{2} S O_{4} \to 5 F e_{2} {(S O_{4})}_{3} + 60 C O_{2} + 60 H N O_{3} + 61 C r_{2} (S O_{4}) + 81 K_{2} S O_{4} + 127 H_{2} O$