A metal oxide has the formula `Z_(2)O_(3)` . It can be reduced by hydrogen to give free metal and water . 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction . The atomic weight of the metal is

Options

A. 27 .9
B. 159.6
C. 79.8
D. 55.8

Explanation

To find the atomic weight of the metal in the oxide \( Z_2O_3 \), we will follow these steps:

### Step 1: Write the Reduction Reaction
The reduction of the metal oxide \( Z_2O_3 \) by hydrogen can be represented as:
\[ Z_2O_3 + 3H_2 \rightarrow 2Z + 3H_2O \]

### Step 2: Determine the Mass of Hydrogen Used
The problem states that 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. We first convert 6 mg to grams:
\[ 6 \text{ mg} = 6 \times 10^{-3} \text{ g} = 0.006 \text{ g} \]

### Step 3: Calculate the Amount of Metal Oxide Reduced by 1 g of Hydrogen
To find out how much \( Z_2O_3 \) can be reduced by 1 g of hydrogen, we set up a proportion:
\[ \text{If } 0.006 \text{ g of H reduces } 0.1596 \text{ g of } Z_2O_3, \]
\[ \text{then } 1 \text{ g of H will reduce } \frac{0.1596 \text{ g}}{0.006 \text{ g}} = 26.6 \text{ g of } Z_2O_3. \]

### Step 4: Calculate the Equivalent Weight of \( Z_2O_3 \)
The equivalent weight of \( Z_2O_3 \) can be calculated using the amount of \( Z_2O_3 \) that can be reduced by 1 g of hydrogen:
\[ \text{Equivalent weight of } Z_2O_3 = 26.6 \text{ g}. \]

### Step 5: Determine the Equivalent Weight of the Metal
The equivalent weight of \( Z_2O_3 \) can be expressed as:
\[ \text{Equivalent weight of } Z_2O_3 = \text{Equivalent weight of } Z + \text{Equivalent weight of } O. \]
Given that the equivalent weight of oxygen (O) is 8 g, we can write:
\[ 26.6 = \text{Equivalent weight of } Z + 8. \]
Thus, the equivalent weight of \( Z \) is:
\[ \text{Equivalent weight of } Z = 26.6 - 8 = 18.6 \text{ g}. \]

### Step 6: Calculate the Atomic Weight of the Metal
The equivalent weight is related to the atomic weight (A) and the valence (n) by the formula:
\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{\text{Valence}}. \]
In \( Z_2O_3 \), the valence of metal \( Z \) is 3. Therefore:
\[ 18.6 = \frac{A}{3}. \]
Solving for \( A \):
\[ A = 18.6 \times 3 = 55.8 \text{ g/mol}. \]

### Conclusion
The atomic weight of the metal \( Z \) is **55.8 g/mol**.

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