Question

We know that balancing of a chemical equation is entirely based on law of conservation of mass. However, the concept of Principle of Atom Conservation (POAC) can also be related to law of conservation of mass in a chemical reaction. So, POAC can also act as a technique for balancing a chemical equation. For example, for a reaction:
$AB{C}_3\to AB+C_2$
On applying POAC for A,B & C and relating the 3 equations, we get : $\frac{n_{AB{C}_3}}{2}=\frac{n_{AB}}{2}=\frac{n_{C_2}}{3}$ ($n_x$: number of moles of X)
Thus, the coefficients of $AB{C}_3$, $AB$ & $C_2$ in the balanced chemical equation will be 2, 2 & 3 respectively and the balanced chemical equation can be represented as:
$2AB{C}_3\to 2AB+3C_2$

Now answer the following question.

Which of the following relation is correct regarding the numerical coefficients p, q, r in the balanced chemical equation, $pA+q{B}_2\to r{A}_2{B}_5$?

• A. $2p=r$
• B. $q=1.25p$
• C. $r=2q$
• D. $q=0.8p$

Answer 1

The correct option is B $q=1.25p$

By applying the concept of Principle of Atom Conservation (POAC) to A and B, following equations can be obtained.

First conserving moles of A, in left hand side of the given equation one A is present and in the right hand side two A is present.

Therefore, $1\times n_A=2\times n_{A_2{B}_5}$ ......(1)

Now conserving the moles of B, in the left hand side of the given equation two B are present and at the right hand side five B are present. Therefore, $2\times n_{B_2}=5\times n_{A_2{B}_5}$ ......(2)

From equations (1) and (2),$\frac{n_A}{4}=\frac{n_{B_2}}{5}=\frac{n_{A_2{B}_5}}{2}$.

Therefore, p is 4, q is 5 and r is 2.

The balanced equation is $4A+5B_2\to 2A_2{B}_5$

Hence, $q=1.25p$.

Option B is correct.