Chemistry Examples

$C r I_{3} + N a (O H) + C l_{2} \to N a I O_{4} + N a_{2} C r O_{4} + N a C l + H_{2} O$

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

$C r$ 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 3

$I$ is not a balanced element. To balance $I$ , multiply $N a I O_{4}$ from the right side by $3$ .

$C r I_{3} + N a (O H) + C l_{2} \to 3 N a I O_{4} + N a_{2} C r O_{4} + N a C l + H_{2} O$

Step 4

$N a$ is not a balanced element. To balance $N a$ , multiply $N a (O H)$ from the left side by $6$ .

$C r I_{3} + 6 N a (O H) + C l_{2} \to 3 N a I O_{4} + N a_{2} C r O_{4} + N a C l + H_{2} O$

Step 5

$O$ is not a balanced element. To balance $O$ , multiply $N a (O H)$ from the left side by $64$ and $3 N a I O_{4}, N a_{2} C r O_{4}, H_{2} O$ from the right side by $2$ .

$C r I_{3} + 64 N a (O H) + C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + N a C l + 2 H_{2} O$

Step 6

$H$ is not a balanced element. To balance $H$ , multiply $2 H_{2} O$ from the right side by $16$ .

$C r I_{3} + 64 N a (O H) + C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + N a C l + 32 H_{2} O$

Step 7

$C l$ is not a balanced element. To balance $C l$ , multiply $N a C l$ from the right side by $2$ .

$C r I_{3} + 64 N a (O H) + C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 2 N a C l + 32 H_{2} O$

Step 8

$C r$ is not a balanced element. To balance $C r$ , multiply $C r I_{3}$ from the left side by $2$ .

$2 C r I_{3} + 64 N a (O H) + C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 2 N a C l + 32 H_{2} O$

Step 9

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

$I$ is a balanced element

Step 10

$N a$ is not a balanced element. To balance $N a$ , multiply $64 N a (O H)$ from the left side by $64$ and $6 N a I O_{4}, 2 N a_{2} C r O_{4}, 2 N a C l$ from the right side by $2$ .

$2 C r I_{3} + 64 N a (O H) + C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 4 N a C l + 32 H_{2} O$

Step 11

$O$ 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 12

$H$ 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 13

$C l$ is not a balanced element. To balance $C l$ , multiply $C l_{2}$ from the left side by $2$ .

$2 C r I_{3} + 64 N a (O H) + 2 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 4 N a C l + 32 H_{2} O$

Step 14

$C r$ 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 15

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

$I$ is a balanced element

Step 16

$N a$ is not a balanced element. To balance $N a$ , multiply $64 N a (O H)$ from the left side by $64$ and $6 N a I O_{4}, 2 N a_{2} C r O_{4}, 4 N a C l$ from the right side by $2$ .

$2 C r I_{3} + 64 N a (O H) + 2 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 8 N a C l + 32 H_{2} O$

Step 17

$O$ 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 18

$H$ 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 19

$C l$ is not a balanced element. To balance $C l$ , multiply $2 C l_{2}$ from the left side by $2$ .

$2 C r I_{3} + 64 N a (O H) + 4 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 8 N a C l + 32 H_{2} O$

Step 20

$C r$ 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 21

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

$I$ is a balanced element

Step 22

$N a$ is not a balanced element. To balance $N a$ , multiply $64 N a (O H)$ from the left side by $64$ and $6 N a I O_{4}, 2 N a_{2} C r O_{4}, 8 N a C l$ from the right side by $2$ .

$2 C r I_{3} + 64 N a (O H) + 4 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 16 N a C l + 32 H_{2} O$

Step 23

$O$ 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 24

$H$ 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 25

$C l$ is not a balanced element. To balance $C l$ , multiply $4 C l_{2}$ from the left side by $2$ .

$2 C r I_{3} + 64 N a (O H) + 8 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 16 N a C l + 32 H_{2} O$

Step 26

$C r$ 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 27

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

$I$ is a balanced element

Step 28

$N a$ is not a balanced element. To balance $N a$ , multiply $64 N a (O H)$ from the left side by $64$ and $6 N a I O_{4}, 2 N a_{2} C r O_{4}, 16 N a C l$ from the right side by $2$ .

$2 C r I_{3} + 64 N a (O H) + 8 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 32 N a C l + 32 H_{2} O$

Step 29

$O$ 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 30

$H$ 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 31

$C l$ is not a balanced element. To balance $C l$ , multiply $8 C l_{2}$ from the left side by $2$ .

$2 C r I_{3} + 64 N a (O H) + 16 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 32 N a C l + 32 H_{2} O$

Step 32

$C r$ 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 33

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

$I$ is a balanced element

Step 34

$N a$ is not a balanced element. To balance $N a$ , multiply $64 N a (O H)$ from the left side by $64$ and $6 N a I O_{4}, 2 N a_{2} C r O_{4}, 32 N a C l$ from the right side by $54$ .

$2 C r I_{3} + 64 N a (O H) + 16 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 54 N a C l + 32 H_{2} O$

Step 35

$O$ 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 36

$H$ 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 37

$C l$ is not a balanced element. To balance $C l$ , multiply $16 C l_{2}$ from the left side by $27$ and from the right side by $16$ .

$2 C r I_{3} + 64 N a (O H) + 27 C l_{2} \to 6 N a I O_{4} + 2 N a_{2} C r O_{4} + 54 N a C l + 32 H_{2} O$