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

Answer

To find the atomic weight of the metal 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 chemical equation:
\[ Z_2O_3 + 3H_2 \rightarrow 2Z + 3H_2O \]
This indicates that 1 mole of \( Z_2O_3 \) reacts with 3 moles of \( H_2 \).

### Step 2: Calculate the number of moles of hydrogen used
We know that 6 mg of hydrogen is used for the reduction. First, we convert this mass into grams:
\[ 6 \text{ mg} = 6 \times 10^{-3} \text{ g} \]
Next, we calculate the number of moles of hydrogen:
\[ \text{Molar mass of } H_2 = 2 \text{ g/mol} \]
\[ \text{Number of moles of } H_2 = \frac{6 \times 10^{-3} \text{ g}}{2 \text{ g/mol}} = 3 \times 10^{-3} \text{ moles} \]

### Step 3: Relate moles of hydrogen to moles of \( Z_2O_3 \)
From the balanced equation, we see that 3 moles of \( H_2 \) react with 1 mole of \( Z_2O_3 \). Therefore, the number of moles of \( Z_2O_3 \) that reacted can be calculated as:
\[ \text{Moles of } Z_2O_3 = \frac{3 \times 10^{-3} \text{ moles of } H_2}{3} = 1 \times 10^{-3} \text{ moles} \]

### Step 4: Calculate the molar mass of \( Z_2O_3 \)
We know the mass of \( Z_2O_3 \) used is 0.1596 g. Using the number of moles calculated, we can find the molar mass \( M \) of \( Z_2O_3 \):
\[ \text{Molar mass of } Z_2O_3 = \frac{\text{mass}}{\text{moles}} = \frac{0.1596 \text{ g}}{1 \times 10^{-3} \text{ moles}} = 159.6 \text{ g/mol} \]

### Step 5: Set up the equation for the molar mass of \( Z_2O_3 \)
The molar mass of \( Z_2O_3 \) can also be expressed in terms of the atomic weight of the metal \( Z \):
\[ \text{Molar mass of } Z_2O_3 = 2X + 3(16) = 2X + 48 \]
Setting this equal to the previously calculated molar mass:
\[ 2X + 48 = 159.6 \]

### Step 6: Solve for \( X \)
Now, we can solve for \( X \):
\[ 2X = 159.6 - 48 = 111.6 \]
\[ X = \frac{111.6}{2} = 55.8 \text{ g/mol} \]

Thus, the atomic weight of the metal \( Z \) is **55.8 g/mol**.