A semiconductor `Y_(2)Ba_(2)Cu_(3)O_(7)` is prepared by a reaction involving `Y_(2)O_(3), BaO_(2)` and CuO. The ratio of their molar should be

Options

A. `1:2:4`
B. `1:2:3`
C. `3:2:1`
D. `1:1.5:2.5`

Answer

To determine the molar ratio of the reactants \( Y_2O_3 \), \( BaO_2 \), and \( CuO \) for the preparation of the semiconductor \( Y_2Ba_2Cu_3O_7 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The first step is to write the balanced chemical equation for the reaction. The reaction can be represented as follows:

\[ Y_2O_3 + BaO_2 + 3CuO \rightarrow Y_2Ba_2Cu_3O_7 + \frac{3}{2}O_2 \]

### Step 2: Identify the coefficients
From the balanced equation, we can identify the coefficients of each reactant:

- \( Y_2O_3 \): 1
- \( BaO_2 \): 2
- \( CuO \): 3

### Step 3: Write the molar ratio
Now that we have the coefficients, we can express the molar ratio of the reactants:

\[ \text{Molar ratio} = 1 : 2 : 3 \]

### Step 4: Conclusion
Thus, the molar ratio of \( Y_2O_3 \), \( BaO_2 \), and \( CuO \) required to prepare \( Y_2Ba_2Cu_3O_7 \) is \( 1 : 2 : 3 \).

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