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

Options

A. 15.58
B. 155.8
C. 5.58
D. 55.8

Answer

To solve the problem, we need to determine the atomic weight of the metal (M) in the metal oxide \( M_2O_3 \) given that 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the reduction reaction
The metal oxide \( M_2O_3 \) can be reduced by hydrogen to form free metal and water. The balanced chemical equation for the reaction is:

\[ M_2O_3 + 3H_2 \rightarrow 2M + 3H_2O \]

### Step 2: Convert the mass of hydrogen to grams
The amount of hydrogen provided is 6 mg. We need to convert this to grams:

\[ 6 \text{ mg} = 6 \times 10^{-3} \text{ g} \]

### Step 3: Determine the amount of metal oxide reduced per gram of hydrogen
From the balanced equation, we see that 1 mole of \( M_2O_3 \) requires 3 moles of \( H_2 \). We need to find out how much \( M_2O_3 \) corresponds to the 6 mg of hydrogen.

Let’s calculate how much \( M_2O_3 \) is reduced by 1 g of hydrogen:

\[ \text{Mass of } M_2O_3 \text{ reduced} = \frac{0.1596 \text{ g}}{6 \times 10^{-3} \text{ g}} = 26.6 \text{ g} \]

### Step 4: Calculate the equivalent weight of the metal oxide
The equivalent weight of \( M_2O_3 \) is 26.6 g. The equivalent weight is defined as:

\[ \text{Equivalent weight} = \text{Equivalent weight of metal} + \text{Equivalent weight of oxygen} \]

### Step 5: Calculate the equivalent weight of oxygen
The atomic weight of oxygen is 16 g. In the oxide \( M_2O_3 \), oxygen has a valency of 2 (as \( O^{2-} \)). Therefore, the equivalent weight of oxygen is:

\[ \text{Equivalent weight of oxygen} = \frac{\text{Atomic weight of oxygen}}{\text{Valency}} = \frac{16}{2} = 8 \text{ g} \]

### Step 6: Calculate the equivalent weight of the metal
Using the equivalent weight of the metal oxide:

\[ \text{Equivalent weight of metal} = \text{Equivalent weight of } M_2O_3 - \text{Equivalent weight of oxygen} = 26.6 \text{ g} - 8 \text{ g} = 18.6 \text{ g} \]

### Step 7: Calculate the atomic weight of the metal
The valency of the metal \( M \) in \( M_2O_3 \) is +3 (as it forms \( M^{3+} \)). Using the formula for equivalent weight:

\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{\text{Valency}} \]

We can rearrange this to find the atomic weight:

\[ \text{Atomic weight} = \text{Equivalent weight} \times \text{Valency} = 18.6 \text{ g} \times 3 = 55.8 \text{ g} \]

### Conclusion
The atomic weight of the metal \( M \) is 55.8 g.