Question

Balance the following chemical equation.
$PbS + {O_2} \to PbO + S{O_2}$

Answer 1

Here, one mole of lead sulphide is reacting with one mole of oxygen and there is a formation of one mole of lead oxide and one mole of sulphur dioxide. According to the law of conservation of mass, the mass of the reactant must be balanced with the mass of the product. Hence, when balancing a chemical equation, the atoms of both molecules as well as elements on the reactant side should be equal to the elements and molecules present in the product side.

Complete answer:
The given reaction is a double displacement reaction. During balancing the equation, the oxygen atom should be kept in an elementary state and we have to balance the other atoms present in the reaction. To balance the number of oxygen atoms, we have to add three in front of the oxygen present on the reactant side. Then the equation will become,
$PbS + 3{O_2} \to PbO + S{O_2}$
Further multiply all the components present in the reactant side and product side with two. Then the balanced equation of the given reaction will become,
$2PbS + 6{O_2} \to 2PbO + 2S{O_2}$
Therefore, the balanced equation can be written as,
$2PbS + 6{O_2} \to 2PbO + 2S{O_2}$

Note:
There are some rules that should be followed when we are balancing the chemical equation. First we have to write the unbalanced equation and start to balance the chemical equation by applying the law of conservation mass. Which means, the same number of atoms should be present in each side of the reaction. Accordingly, we have to add the elements until the number of molecules is equal on both LHS and RHS.