A metal oxide has the formul `Z_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water. `0.2 g` of the metal oxide requires `12 mg` of hydrogen for complete reduction. The atomic weight of the metal is

Options

A. 52
B. 104
C. 26
D. 78

Explanation

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

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

### Step 2: Calculate Moles of Metal Oxide
Given that the mass of the metal oxide \( Z_2O_3 \) is \( 0.2 \, \text{g} \), we can express the number of moles of \( Z_2O_3 \) using its molar mass \( M \):
\[ \text{Number of moles of } Z_2O_3 = \frac{0.2}{M} \]

### Step 3: Convert Mass of Hydrogen to Grams
The mass of hydrogen used for the reduction is given as \( 12 \, \text{mg} \). We convert this to grams:
\[ 12 \, \text{mg} = \frac{12}{1000} \, \text{g} = 0.012 \, \text{g} \]

### Step 4: Calculate Moles of Hydrogen
The molar mass of hydrogen \( H_2 \) is \( 2 \, \text{g/mol} \). Therefore, the number of moles of hydrogen is:
\[ \text{Number of moles of } H_2 = \frac{0.012}{2} = 0.006 \, \text{moles} \]

### Step 5: Relate Moles of Hydrogen to Moles of Metal Oxide
From the balanced equation, we see that 3 moles of \( H_2 \) react with 1 mole of \( Z_2O_3 \). Thus, the moles of \( Z_2O_3 \) that react with \( 0.006 \) moles of \( H_2 \) can be calculated as:
\[ \text{Moles of } Z_2O_3 = \frac{0.006}{3} = 0.002 \, \text{moles} \]

### Step 6: Set Up the Equation
Now we can set the number of moles of \( Z_2O_3 \) from the mass and the calculated moles:
\[ \frac{0.2}{M} = 0.002 \]

### Step 7: Solve for Molar Mass \( M \)
Rearranging the equation to solve for \( M \):
\[ M = \frac{0.2}{0.002} = 100 \, \text{g/mol} \]

### Step 8: Calculate Atomic Mass of Metal \( Z \)
The molar mass of \( Z_2O_3 \) can be expressed in terms of the atomic mass of \( Z \) (let's denote it as \( x \)):
\[ 2x + 3(16) = 100 \]
\[ 2x + 48 = 100 \]
\[ 2x = 100 - 48 = 52 \]
\[ x = \frac{52}{2} = 26 \, \text{g/mol} \]

### Conclusion
The atomic weight of the metal \( Z \) is \( 26 \, \text{g/mol} \).

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