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)`
If in the above question, the atomic masses of X and Y are 10 and 30 respectively, then the mass of `XY_(3)` formed when 120 g of `Y_(2)` reacts completely with X is:

Options

A. 133.3 g
B. 200 g
C. 266.5 g
D. 400 g.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
From the information provided, we can represent the reaction as:
\[ 2X + 3Y_2 \rightarrow 2XY_3 \]

### Step 2: Determine the molar masses
- The atomic mass of \( X \) is given as 10 g/mol.
- The atomic mass of \( Y \) is given as 30 g/mol.
- Therefore, the molar mass of \( Y_2 \) (which contains 2 atoms of \( Y \)) is:
\[ \text{Molar mass of } Y_2 = 2 \times 30 = 60 \text{ g/mol} \]
- The molar mass of \( XY_3 \) (which contains 1 atom of \( X \) and 3 atoms of \( Y \)) is:
\[ \text{Molar mass of } XY_3 = 10 + (3 \times 30) = 10 + 90 = 100 \text{ g/mol} \]

### Step 3: Calculate the moles of \( Y_2 \)
Given that the mass of \( Y_2 \) is 120 g, we can calculate the number of moles of \( Y_2 \):
\[ \text{Moles of } Y_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{120 \text{ g}}{60 \text{ g/mol}} = 2 \text{ moles} \]

### Step 4: Use stoichiometry to find moles of \( XY_3 \) produced
From the balanced equation, we see that 3 moles of \( Y_2 \) produce 2 moles of \( XY_3 \). Therefore, we can set up a ratio:
\[ \frac{2 \text{ moles of } XY_3}{3 \text{ moles of } Y_2} = \frac{x \text{ moles of } XY_3}{2 \text{ moles of } Y_2} \]
Cross-multiplying gives:
\[ x = \frac{2 \times 2}{3} = \frac{4}{3} \text{ moles of } XY_3 \approx 1.33 \text{ moles of } XY_3 \]

### Step 5: Calculate the mass of \( XY_3 \) produced
Now, we can calculate the mass of \( XY_3 \) produced using the number of moles:
\[ \text{Mass of } XY_3 = \text{Moles} \times \text{Molar mass} = 1.33 \text{ moles} \times 100 \text{ g/mol} = 133.33 \text{ g} \]

### Final Answer
The mass of \( XY_3 \) formed when 120 g of \( Y_2 \) reacts completely with \( X \) is approximately **133.33 g**.

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