For many school students to write the equations of chemical reactions and it is correct to place coefficients a hard task. And, the main difficulty in them for some reason is caused by its second part. It would seem, there is nothing difficult in that, however sometimes pupils give in, falling into full confusion. And it is necessary to remember only several simple rules, and the task will cease to cause difficulties.
Instruction
1. The coefficient, that is the number facing a chemical molecule formula belongs to all symbols, and is multiplied by each index of each symbol! It is multiplied, but it does not develop! It can seem improbable, but some school students put two numbers instead of to multiply them.
2. In other words, if in the left part of reaction it is written down: 2Na3PO4 + 3CaCl2 = … It means that 6 atoms of sodium, 2 atoms of phosphorus, 8 atoms of oxygen, 3 atoms of calcium and 6 atoms of chlorine reacted.
3. The amount of atoms of each element of initial substances (that is being in the left member of equation) has to coincide with amount of atoms of each element of products of reaction (respectively, being in its right part).
4. Let's consider this rule, having written down up to the end the sodium phosphate reaction equation with chloride calcium. For descriptive reasons, remove coefficients from the left member of equation. Na3PO4 + CaCl2 = Ca3(PO4) 2 + NaCl
5. During reaction almost insoluble salt – calcium phosphate – and chloride sodium is formed. How to place coefficients? At once pay attention that at phosphate ion (PO4) the index is equal in the right member of equation to two. Therefore to balance amounts of atoms of phosphorus and oxygen in the left and right part, before a formula of a molecule of phosphate of sodium it is necessary to deliver coefficient 2. It will turn out: 2Na3PO4 + CaCl2 = Ca3(PO4) 2 + NaCl
6. You see that the amount of atoms of phosphorus and oxygen is balanced, but still variously amount of atoms of sodium, calcium and chlorine. In the left part: sodium – 6 atoms, calcium – 1 atom, chlorine – 2 atoms. In the right part, respectively: sodium – 1 atom, calcium – 3 atom, chlorine – 1 atom.
7. Equalize amount of atoms of sodium, having attributed coefficient 6 to a molecule of chloride sodium. It turns out: 2Na3(PO4) 2 + CaCl2 = Ca3(PO4)2 + 6NaCl
8. It was necessary to balance the last two elements. You see that in the left part 1 atoms of calcium and 2 atoms of chlorine, and in right - 3 atoms of calcium and 6 atoms of chlorine. That is, exactly three times more! Substituting coefficient 3 to a molecule of chloride calcium, receive the total equation: 2Na3(PO4) 2 + 3CaCl2 = Ca3(PO4)2 + 6NaCl