Translate the following statements into chemical equations and then balance them:
(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal react with water to give potassium hydroxide and hydrogen gas.

Explanation

(a) The symbol equation for the reaction is :
`H_(2) + N_(2) to NH_(3)`
The balancing of equation is done in the following steps :
Step I. Let us count the number of atoms of all the elements fo the reactants and the products on both sides of the equation.

A simple look at the equation reveals that neither the number of H nor of N atoms are equal on both sides of the equation.
Step II. In order to equate the number of H atoms on both sides, put coefficient 3 before `H_(2)` on the reactant side and coefficient 2 before `NH_(3)` on the product side.
`3H_(2) + N_(2) to 2NH_(3)`
Step III. On counting, the number of N atoms on both sides of the equation are also the same (2). This means that the equation is balanced.
(b) The symbol equation for the reaction is :
`H_(2)S + O_(2) to H_(2)O + SO_(2)`
The balancing of equation is done in the following steps :
Step I. Let us count the number of atoms of all the elements on both sides on the equation.

A simple look at the equation reveals that the number of H and S atoms are equal on both sides. At the same time, the number of O atoms are not equal.
Step II. In order to equate the number of O atoms, put coefficient 3 before `O_(2)` on the reactant side and coefficient 2 before `SO_(2)` on the product side.
`H_(2)S + 3O_(2) to H_(2)O + 2SO_(2)`
Step III. O atoms are still not balanced. To achieve this, put coefficient 2 before `H_(2)O` on the product side.
`H_(2)S + 3O_(2) to 2H_(2)O + 2SO_(2)`
Step IV. To balance S atoms, put coefficient 2 before `H_(2)S` on the reactant side.
`2H_(2)S + 3O_(2) to 2H_(2)O + 2SO_(2)`
Step V. On inspection, the number of atoms of all the elements in both sides of the equation are equal. Therefore, the equation is balanced.
(c) The symbol equation if done in the following steps :
Step I. Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that only Ba atoms are equal on both sides. The rest of the atoms are to be balanced. It is done as follows :
Step II. In order to equate the number of Al atoms, put coefficient 2 before `AlCl_(3)` on the product side `BaCl_(2) + Al_(2)(SO_(4))_(3) to 2AlCl_(3) + BaSO_(4)`
Step III. In order to balance Cl atoms, put coefficient 3 before `BaCl_(2)` on the reactant side.
`3BaCl_(2) + Al_(2)(SO_(4))_(3) to 2AlCl_(3) + BaSO_(4)`
Step IV. To balance Ba atoms, put coefficient 3 before `BaSO_(4)` on the product side.
`3BaCl_(2) + Al_(2)(SO_(4))_(3) to 2AlCl_(3) + 3BaSO_(4)`
Step V. On inspection, the number of S and O atoms on both sides of the equation are also found to be equal. Thus, the equation is in balanced form.
(d) The symbol equation for the reaction is :
`K + H_(2)O to KOH + H_(2)`
The balancing of the equation is done in the following steps :
Step I. Let us count the number of atoms of all the elements on both sides.

A simple look at the equation reveals that the number fo K and O atoms on the both sides of the equation are equal. At the same time, the number of H atoms are not equal.
Step II. To balance the number of H atoms, put coefficient 2 before KOH on the product side and 2 before `H_(2)O` on the reactant side.
`K + 2H_(2)O to 2KOH + H_(2)`
Step III. To balance the number of K atoms in the above equation, put coefficient 2 before K atom on the reactant side.
`2K + 2H_(2)O to 2KOH + H_(2)`
Step IV. On inspection, the number of atoms of all the elements are found to be equal on both sides of the equation. It is balanced.