Consider the following chemical equation :

$pAl + qH_2O\longrightarrow rAl_2O_3 + sH_2$

To balance this chemical equation, the values of 'p', 'q', 'r' and 's' must be respectively :

The skeletal equation is $pAl + qH_2O \longrightarrow rAl_2O_3 + sH_2$.

As given in the NCERT method of balancing chemical equations (Section 1.1.2), we box each formula and balance the atom with the maximum count first — here it is convenient to start with $Al_2O_3$.

Balancing Al: With $r=1$ (i.e., one $Al_2O_3$), 2 Al atoms are needed on RHS, so $p=2$:

$$2Al + qH_2O \longrightarrow Al_2O_3 + sH_2$$

Balancing O: $Al_2O_3$ has 3 oxygen atoms, and each $H_2O$ has 1 oxygen atom, so $q=3$:

$$2Al + 3H_2O \longrightarrow Al_2O_3 + sH_2$$

Balancing H: LHS has $3\times2=6$ H atoms (from $3H_2O$); RHS needs $s=3$ (since each $H_2$ has 2 atoms, $3\times2=6$):

$$2Al + 3H_2O \longrightarrow Al_2O_3 + 3H_2$$

Check (as per the law of conservation of mass, Section 1.1.2):

All atoms match on both sides, so the balanced equation is $2Al + 3H_2O \rightarrow Al_2O_3 + 3H_2$, giving $p,q,r,s = 2,3,1,3$.

Option (A) 3, 2, 2, 1 — incorrect: substituting gives $3Al + 2H_2O \rightarrow 2Al_2O_3 + H_2$; Al count is 3 on LHS but 4 on RHS, and O count is 2 on LHS but 6 on RHS — mass is not conserved.

Option (B) 2, 3, 3, 1 — incorrect: gives $2Al + 3H_2O \rightarrow 3Al_2O_3 + H_2$; Al is 2 on LHS but 6 on RHS, and O is 3 on LHS but 9 on RHS — unbalanced.

Option (D) 3, 1, 2, 2 — incorrect: gives $3Al + H_2O \rightarrow 2Al_2O_3 + 2H_2$; Al is 3 on LHS but 4 on RHS, and O is 1 on LHS but 6 on RHS — unbalanced.

Hence only option (C) satisfies the requirement (stated in Section 1.1.2) that the number of atoms of each element must be the same on both sides of a chemical equation.