Question

Complete the following reactions:
\[PbS+4{{O}_{3}}\to PbS{{O}_{4}}+\_\_\_\_\_\_\]
\[Cu+2{{H}_{2}}S{{O}_{4}}\to CuS{{O}_{4}}+\_\_\_\_\_\_+2{{H}_{2}}O\]
\[C{{l}_{2}}+2{{H}_{2}}O+S{{O}_{2}}\to \_\_\_\_\_\_\_+2HCl\]

Answer 1

Hint: Since, the reactions are already balanced and the stoichiometric coefficients are given, calculate the atoms that are present on the right-hand side and formulate the most probable and stable molecule that can be formed there.

Complete answer:
a. \[PbS+4{{O}_{3}}\to PbS{{O}_{4}}+\_\_\_\_\_\_\]
Consider the number of atoms present on both sides:

Atom	LHS	RHS
Pb	1	1
S	1	1
O	12	4

Thus, the right-hand side has a deficit of 8 O atoms.
The ozone molecule on the left-hand side is highly unstable and will degrade to an oxygen molecule whenever possible. The blank can be filled up with 4 molecules of oxygen to cover up the deficit.
1mol of lead sulphide will combine with 4mol of ozone to form 1mol of lead sulphate and 4mol of oxygen.
\[PbS+4{{O}_{3}}\to PbS{{O}_{4}}+4{{O}_{2}}\]
b. \[Cu+2{{H}_{2}}S{{O}_{4}}\to CuS{{O}_{4}}+\_\_\_\_\_\_+2{{H}_{2}}O\]
Consider the number of atoms present on both sides:

Atom	LHS	RHS
Cu	1	1
S	2	1
O	8	6
H	4	4

Thus, the right-hand side has a deficit of 1 S atom and 2 O atoms
As we know, copper is usually unreactive, but when the copper plate is heated and concentrated sulphuric acid is poured on it, a gas is liberated, this gas is the toxic sulphur dioxide. Thus, the blank can be filled up with 1 molecule of sulphur dioxide to cover up the deficit.
1mol of heated copper will react with 2mol of concentrated sulphuric acid to give 1mol of copper sulphate, 1mol of sulphur dioxide and 2mol of water.
\[Cu+2{{H}_{2}}S{{O}_{4}}\to CuS{{O}_{4}}+S{{O}_{2}}+2{{H}_{2}}O\]
c. \[C{{l}_{2}}+2{{H}_{2}}O+S{{O}_{2}}\to \_\_\_\_\_\_\_+2HCl\]
Consider the number of atoms present on both sides:

Atom	LHS	RHS
Cl	2	2
S	1	0
O	4	0
H	4	2

Thus, the right-hand side has a deficit of 2 H atoms, 1 S atom, and 4 O atoms.
Here, sulphur dioxide reacts with chlorine in the presence of moisture to form 2 acidic compounds. One of them is the hydrochloric acid and given the deficit of atoms, we can fill the blank with 1 molecule of sulphuric acid.
1mol of chlorine gas will react with 1mol of sulphur dioxide in the presence of 2mol of water to give 1mol of sulphuric acid and 2mol of hydrochloric acid.
\[C{{l}_{2}}+2{{H}_{2}}O+S{{O}_{2}}\to {{H}_{2}}S{{O}_{4}}+2HCl\]

Note: Always check if the reaction is balanced or not before attempting to formulate the unknown products. This will throw off the entire reaction. Here, in the second example, copper has to be in the presence of 2 moles of sulphuric acid and has to be heated to produce sulphur dioxide and water as byproducts. IF these conditions are not met, it will form hydrogen gas as the byproduct.