`"2.6 g of CaCO"_(3) and MgCO_(3)` mixture is heated to a constant weight of 1.3 g. The mass of `CaCO_(3)` in mixture is :

Options

A. 980 mg
B. 400 mg
C. 1.75 g
D. 0.74 g

Answer

To find the mass of CaCO₃ in the mixture of CaCO₃ and MgCO₃ that weighs 2.6 g and upon heating reduces to 1.3 g, we can follow these steps:

### Step 1: Define Variables
Let the mass of CaCO₃ in the mixture be \( x \) grams. Therefore, the mass of MgCO₃ in the mixture will be \( 2.6 - x \) grams.

### Step 2: Calculate the Mass Loss
When heated, CaCO₃ decomposes into CaO and CO₂, and MgCO₃ decomposes into MgO and CO₂. The total mass loss during the heating process is given by:

\[ \text{Initial mass} - \text{Final mass} = 2.6 \, \text{g} - 1.3 \, \text{g} = 1.3 \, \text{g} \]

### Step 3: Determine the Mass of CaO Produced
The molar mass of CaCO₃ is 100 g/mol, and it produces 56 g of CaO. Therefore, the mass of CaO produced from \( x \) grams of CaCO₃ can be calculated using the unitary method:

\[ \text{Mass of CaO from } x \, \text{g of CaCO₃} = \frac{56}{100} \times x = 0.56x \, \text{g} \]

### Step 4: Determine the Mass of MgO Produced
The molar mass of MgCO₃ is 84 g/mol, and it produces 40 g of MgO. Therefore, the mass of MgO produced from \( 2.6 - x \) grams of MgCO₃ can be calculated as:

\[ \text{Mass of MgO from } (2.6 - x) \, \text{g of MgCO₃} = \frac{40}{84} \times (2.6 - x) = \frac{40}{84} (2.6 - x) \, \text{g} \]

### Step 5: Set Up the Equation
The total mass of CaO and MgO produced must equal the mass loss of 1.3 g:

\[ 0.56x + \frac{40}{84}(2.6 - x) = 1.3 \]

### Step 6: Simplify the Equation
Multiply through by 84 to eliminate the fraction:

\[ 84 \times 0.56x + 40(2.6 - x) = 1.3 \times 84 \]

Calculating each term gives:

\[ 47.04x + 104 - 40x = 109.2 \]

### Step 7: Combine Like Terms
Combine the \( x \) terms:

\[ 7.04x + 104 = 109.2 \]

### Step 8: Isolate \( x \)
Subtract 104 from both sides:

\[ 7.04x = 5.2 \]

Now, divide by 7.04:

\[ x = \frac{5.2}{7.04} \approx 0.738 \, \text{g} \]

### Step 9: Round the Result
Rounding to two decimal places gives:

\[ x \approx 0.74 \, \text{g} \]

### Conclusion
The mass of CaCO₃ in the mixture is approximately **0.74 grams**.

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