A metal carbonate decomposes according to the following reaction
`M_(2)CO_(3)` (s) `to` `M_(2)O`(s) + `CO_(2)`(g)
Percentage loss in mass on complete decomposition of `M_(2)CO_(3)`(s)
(Atomic mass of M= 102 )

Options

A. `100/3 %`
B. `50/3 %`
C. `25/3 %`
D. `15%`

Answer

To solve the problem of finding the percentage loss in mass on the complete decomposition of \( M_2CO_3 \), we will follow these steps:

### Step 1: Write the decomposition reaction
The decomposition reaction of the metal carbonate is given as:
\[ M_2CO_3 (s) \rightarrow M_2O (s) + CO_2 (g) \]

### Step 2: Calculate the molar mass of \( M_2CO_3 \)
The molar mass of \( M_2CO_3 \) can be calculated using the atomic mass of M (given as 102 g/mol):
- Molar mass of \( M_2CO_3 \) = \( 2 \times \text{Atomic mass of M} + \text{Atomic mass of C} + 3 \times \text{Atomic mass of O} \)
- Molar mass of \( M_2CO_3 \) = \( 2 \times 102 + 12 + 3 \times 16 \)
- Molar mass of \( M_2CO_3 \) = \( 204 + 12 + 48 = 264 \, \text{g/mol} \)

### Step 3: Calculate the molar mass of \( M_2O \)
The molar mass of \( M_2O \) is calculated as follows:
- Molar mass of \( M_2O \) = \( 2 \times \text{Atomic mass of M} + \text{Atomic mass of O} \)
- Molar mass of \( M_2O \) = \( 2 \times 102 + 16 \)
- Molar mass of \( M_2O \) = \( 204 + 16 = 220 \, \text{g/mol} \)

### Step 4: Calculate the molar mass of \( CO_2 \)
The molar mass of \( CO_2 \) is calculated as:
- Molar mass of \( CO_2 \) = \( \text{Atomic mass of C} + 2 \times \text{Atomic mass of O} \)
- Molar mass of \( CO_2 \) = \( 12 + 2 \times 16 \)
- Molar mass of \( CO_2 \) = \( 12 + 32 = 44 \, \text{g/mol} \)

### Step 5: Calculate the mass loss during the reaction
The mass loss during the decomposition reaction is equal to the mass of \( CO_2 \) produced:
- Mass loss = Molar mass of \( CO_2 \) = 44 g

### Step 6: Calculate the percentage loss in mass
The percentage loss in mass can be calculated using the formula:
\[ \text{Percentage loss} = \left( \frac{\text{Mass loss}}{\text{Initial mass}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage loss} = \left( \frac{44}{264} \right) \times 100 = \frac{44 \times 100}{264} = \frac{4400}{264} \approx 16.67\% \]

### Step 7: Final answer
Thus, the percentage loss in mass on complete decomposition of \( M_2CO_3 \) is:
\[ \text{Percentage loss} \approx 16.67\% \]