Transfer the following into chemical equations and 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 precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
(a) The symbol equation for the reaction is:
H2 + N2 ➝ NH3
The balancing of equation is done in the following steps:
Step I: Let us count the number of atoms of all the elements of the reactants and the products on both sides of the equation.
Elements : H
No. of atoms of reactants (LHS): 2
No. of atoms of products (RHS) : 3
Elements : N
No. of atoms of reactants (LHS): 2
No. of atoms of products (RHS) : 1
Step II: In order to equate the number of H atoms on both sides, put coefficient 3 before H2 on the reactant side and coefficient 2 before NH3 on the product side.
3H2 + N2 ➝ 2NH3
Step III: On counting, the number of N atoms on both sides of the equation are same. This means that the equation is balanced.
Step IV: The balanced equation can be written in physical state as:
3H2(g) + N2(g) ➝ 2NH3(g)
(b) The symbolic equation for the reaction is:
H2S + O2 ➝ H2O + SO2
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 of the equation.
Elements : H
No. of atoms of reactants (LHS): 2
No. of atoms of products (RHS) : 2
Elements : S
No. of atoms of reactants (LHS): 1
No. of atoms of products (RHS) : 1
Elements : O
No. of atoms of reactants (LHS): 2
No. of atoms of products (RHS) : 3
Step II: In order to equate the number of O atoms, put coefficient 3 before O2 on the reactant side and coefficient 2 before SO2 on the product side.
H2S + 3O2 ➝ H2O + 2SO2
Step III: 0 atoms are still not balanced. To achieve this, put coefficient 2 before H2O on the product side.
H2S + 3O2 ➝ 2H2O + 2S02
Step IV: To balance S atoms, put coefficient 2 before H2S on the reactant side.
2H2S + 3O2 ➝ 2H2O + 2SO2
Step V: On inspection, the number of atoms of all the elements on both sides of the equation are equal. Therefore, the equation is balanced.
Step VI: The balanced equation can be written in physical state as:
2H2S(g) + 3O2(g) ➝ 2H2O(g) + 2SO2(g)
(c) The symbolic equation for the reaction is:
BaCl2 + Al2(SO4)3 ➝ AlCl3 + BaSO4
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 of the equation.
Elements : Ba
No. of atoms of reactants (LHS): 1
No. of atoms of products (RHS) : 1
Elements : Al
No. of atoms of reactants (LHS): 2
No. of atoms of products (RHS) : 1
Elements : Cl
No. of atoms of reactants (LHS): 2
No. of atoms of products (RHS) : 3
Elements : S
No. of atoms of reactants (LHS): 3
No. of atoms of products (RHS) : 1
Elements : O
No. of atoms of reactants (LHS): 12
No. of atoms of products (RHS) : 4
(d) The symbolic equation for the reaction is:
K + H2O ➝ KOH + H2
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.
Elements : K
No. of atoms of reactants (LHS): 1
No. of atoms of products (RHS) : 1
Elements : H
No. of atoms of reactants (LHS): 2
No. of atoms of products (RHS) : 3
Elements : O
No. of atoms of reactants (LHS): 1
No. of atoms of products (RHS) : 1
Step II: To balance the number of H atoms, put coefficient 2 before KOH on the product side and 2 before H2O on the reactant side.
K + 2H2O ➝ 2KOH + H2
Step III: To balance the number of K atoms in the above equation, put coefficient 2 before K atom on the reactant side.
2K + 2H2O ➝ 2KOH + H2
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 finally balanced.
Step V: The balanced equation can be written in physical state as:
2K(s) + 2H2O(l) ➝ 2KOH(aq) + H2(g)