12 grams of a mixture of sand and calcium carbonate on strong heating produced 7.6 grams of residue. How many grams of sand is present in the mixture?

Answer

To solve the problem of how many grams of sand are present in a mixture of sand and calcium carbonate, we can follow these steps:

### Step 1: Define Variables
Let the mass of sand in the mixture be \( x \) grams. Therefore, the mass of calcium carbonate in the mixture will be:
\[ \text{Mass of calcium carbonate} = 12 - x \text{ grams} \]

### Step 2: Understand the Reaction
When the mixture is heated, calcium carbonate decomposes into calcium oxide and carbon dioxide:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
Sand does not decompose and remains as is. The residue after heating consists of sand and calcium oxide.

### Step 3: Calculate the Mass of Calcium Oxide Produced
From the stoichiometry of the reaction, we know that:
- 1 mole of calcium carbonate (100 g) produces 1 mole of calcium oxide (56 g).

Thus, the mass of calcium oxide produced from \( 12 - x \) grams of calcium carbonate can be calculated as:
\[ \text{Mass of calcium oxide} = \left(\frac{56}{100}\right) \times (12 - x) = 0.56(12 - x) \text{ grams} \]

### Step 4: Set Up the Equation for the Residue
The total mass of the residue after heating is given as 7.6 grams. The residue consists of the mass of sand and the mass of calcium oxide:
\[ x + 0.56(12 - x) = 7.6 \]

### Step 5: Simplify and Solve the Equation
Expanding the equation:
\[ x + 6.72 - 0.56x = 7.6 \]
Combining like terms:
\[ (1 - 0.56)x + 6.72 = 7.6 \]
\[ 0.44x + 6.72 = 7.6 \]
Subtracting 6.72 from both sides:
\[ 0.44x = 7.6 - 6.72 \]
\[ 0.44x = 0.88 \]
Dividing both sides by 0.44:
\[ x = \frac{0.88}{0.44} = 2 \]

### Step 6: Conclusion
The mass of sand present in the mixture is:
\[ \text{Mass of sand} = 2 \text{ grams} \]