Equal weights of `X` (atomic weight = 36) and `Y` (atomic weight = 24) are reacted to form the compound `X_(2) Y_(3)`, which of the following is/are correct

Options

A. `X` is the limiting reagent
B. `Y` is the limiting reagen.
C. NO reactant is left over.
D. Mass of `X_(2)Y_(3)` formed is double the mass of `X` taken.

Explanation

To solve the problem, we need to analyze the reaction between equal weights of elements X and Y, and determine the properties of the compound formed, X₂Y₃.

### Step-by-Step Solution:

1. **Identify the Atomic Weights**:
- Atomic weight of X = 36 g/mol
- Atomic weight of Y = 24 g/mol

2. **Assume Equal Weights**:
- Let the equal weight of X and Y taken for the reaction be 'a' grams.

3. **Calculate the Initial Moles**:
- Moles of X = \( \frac{a}{36} \)
- Moles of Y = \( \frac{a}{24} \)

4. **Write the Balanced Chemical Equation**:
- The balanced reaction is:
\[ 2X + 3Y \rightarrow X_2Y_3 \]

5. **Determine the Stoichiometry**:
- According to the balanced equation, 2 moles of X react with 3 moles of Y.
- From the moles calculated:
- For X: \( \frac{a}{36} \) moles
- For Y: \( \frac{a}{24} \) moles

6. **Find the Limiting Reagent**:
- To find which reactant is limiting, we can set up a ratio based on the stoichiometry:
- Required moles of Y for the available moles of X:
\[ \text{Required moles of Y} = \frac{3}{2} \times \frac{a}{36} = \frac{3a}{72} \]
- Compare this with the available moles of Y:
\[ \frac{a}{24} \]
- Since \( \frac{3a}{72} = \frac{a}{24} \), both reactants are consumed completely, indicating that there is no limiting reagent.

7. **Calculate the Moles of the Product (X₂Y₃)**:
- The moles of X₂Y₃ formed can be calculated as:
\[ \text{Moles of } X_2Y_3 = \frac{a}{72} \]

8. **Calculate the Molar Mass of X₂Y₃**:
- Molar mass of X₂Y₃:
\[ \text{Molar mass} = 2 \times 36 + 3 \times 24 = 72 + 72 = 144 \text{ g/mol} \]

9. **Calculate the Weight of X₂Y₃ Formed**:
- Weight of X₂Y₃ = Moles × Molar Mass:
\[ \text{Weight of } X_2Y_3 = \left(\frac{a}{72}\right) \times 144 = 2a \]

10. **Conclusion**:
- The mass of X₂Y₃ formed is double the mass of Y taken (since Y was taken as 'a' grams).
- Therefore, the statements regarding the limiting reagent and the mass of the product formed can be evaluated based on this analysis.

### Final Answers:
- The correct options are:
- The mass of X₂Y₃ formed is double the mass of Y taken.
- There is no limiting reagent.