18.4 gram mixture of `MgCO_(3)` and `CaCO_(3)` produce 4.48 litre of `CO_(2)` at STP. Then calculate the amount of `MgCO_(3)` and `CaCO_(3)` in mixture.

Answer

To solve the problem of finding the amounts of magnesium carbonate (MgCO₃) and calcium carbonate (CaCO₃) in an 18.4 gram mixture that produces 4.48 liters of CO₂ at STP, we can follow these steps:

### Step 1: Calculate the moles of CO₂ produced
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the moles of CO₂ produced from the given volume.

\[ \text{Moles of CO₂} = \frac{\text{Volume of CO₂}}{22.4 \text{ L/mol}} = \frac{4.48 \text{ L}}{22.4 \text{ L/mol}} = 0.2 \text{ moles} \]

### Step 2: Calculate the mass of CO₂ produced
The molar mass of CO₂ is calculated as follows:
- Molar mass of C = 12 g/mol
- Molar mass of O = 16 g/mol
- Therefore, molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol

Now, we can calculate the mass of CO₂ produced:

\[ \text{Mass of CO₂} = \text{Moles of CO₂} \times \text{Molar mass of CO₂} = 0.2 \text{ moles} \times 44 \text{ g/mol} = 8.8 \text{ g} \]

### Step 3: Set up equations for the mixture
Let the mass of CaCO₃ in the mixture be \( x \) grams. Then, the mass of MgCO₃ will be \( 18.4 - x \) grams.

### Step 4: Write the equations based on the decomposition reactions
1. For CaCO₃:
- 100 g of CaCO₃ produces 44 g of CO₂.
- Therefore, \( x \) grams of CaCO₃ produces \( \frac{44}{100} \times x \) grams of CO₂.

2. For MgCO₃:
- 84 g of MgCO₃ produces 44 g of CO₂.
- Therefore, \( 18.4 - x \) grams of MgCO₃ produces \( \frac{44}{84} \times (18.4 - x) \) grams of CO₂.

### Step 5: Set up the total CO₂ equation
The total mass of CO₂ produced from both reactions equals 8.8 g:

\[ \frac{44}{100}x + \frac{44}{84}(18.4 - x) = 8.8 \]

### Step 6: Solve the equation
Multiply through by 4200 (the least common multiple of 100 and 84) to eliminate the fractions:

\[ 44 \times 42x + 44 \times 50(18.4 - x) = 8.8 \times 4200 \]

This simplifies to:

\[ 1848x + 2200(18.4 - x) = 36960 \]

Expanding the equation gives:

\[ 1848x + 40480 - 2200x = 36960 \]

Combining like terms:

\[ -352x + 40480 = 36960 \]

Rearranging gives:

\[ -352x = 36960 - 40480 \]
\[ -352x = -3520 \]
\[ x = 10 \]

### Step 7: Calculate the masses of each component
Now that we have \( x = 10 \) grams for CaCO₃, we can find the mass of MgCO₃:

\[ \text{Mass of MgCO₃} = 18.4 - x = 18.4 - 10 = 8.4 \text{ grams} \]

### Final Answer
- Mass of CaCO₃ = 10 grams
- Mass of MgCO₃ = 8.4 grams