We know that balancing of a chemical equation is entirely bases on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to law of consevation of mass in a chemical reaction. So, POAC can also act as a technique for balancing a chemical equation. For example, for a reaction:
`ABC_(3) rarr AB+C_(2)`
On applying POAC for A , B and C and related the equations , we get : `(n_(ABC_(3)))/(2) =(n_(AB))/(2) =(n_(C_(2)))/(3)`
(`n_(x)` : number of moles of X)
Thus , the cofficients of `ABC_(3)` , AB and `C_(2)` in the balanced chemical equation will be 2,2 and 3 respectively and the balanced chemical equation can be represented as ,
`2ABC_(3) rarr 2AB + 3C_(2)` Which of the following statements jis correct balanced chemical equation,
`pA+qB_(2) rarr rA_(2)B_(5)`

Options

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

Explanation

To solve the problem of balancing the chemical equation \( pA + qB_2 \rightarrow rA_2B_5 \) using the Principle of Atom Conservation (POAC), we will follow these steps:

### Step 1: Write down the unbalanced equation
The unbalanced equation is:
\[ pA + qB_2 \rightarrow rA_2B_5 \]

### Step 2: Identify the number of atoms of each element on both sides
- On the left side:
- A: \( p \)
- B: \( 2q \) (since \( B_2 \) has 2 atoms of B)

- On the right side:
- A: \( 2r \) (since \( A_2 \) has 2 atoms of A)
- B: \( 5r \) (since \( B_5 \) has 5 atoms of B)

### Step 3: Set up equations based on the conservation of atoms
From the above counts, we can set up the following equations:

1. For element A:
\[ p = 2r \quad \text{(1)} \]

2. For element B:
\[ 2q = 5r \quad \text{(2)} \]

### Step 4: Solve the equations for p, q, and r
From equation (1):
\[ p = 2r \]

From equation (2):
\[ q = \frac{5r}{2} \]

### Step 5: Assign a value to r
To find the simplest integer values for \( p \), \( q \), and \( r \), we can assign \( r = 2 \):
- Substituting \( r = 2 \) into the equations:
- From (1):
\[ p = 2 \times 2 = 4 \]
- From (2):
\[ q = \frac{5 \times 2}{2} = 5 \]

### Step 6: Write the balanced equation
Now substituting the values of \( p \), \( q \), and \( r \) into the original equation:
\[ 4A + 5B_2 \rightarrow 2A_2B_5 \]

### Conclusion
The balanced chemical equation is:
\[ 4A + 5B_2 \rightarrow 2A_2B_5 \]

### Step 7: Find the relation between p, q, and r
From the values we found:
- \( p = 4 \)
- \( q = 5 \)
- \( r = 2 \)

We can check the relations:
1. \( 2p = r \) → \( 2 \times 4 \neq 2 \) (not valid)
2. \( q = 1.25p \) → \( 5 = 1.25 \times 4 \) (valid)
3. \( r = 2^3 \) → \( 2 \neq 8 \) (not valid)
4. \( q = 0.8p \) → \( 5 \neq 0.8 \times 4 \) (not valid)

Thus, the correct relation is:
\[ \text{Option B: } q = 1.25p \]