What mass of hydrochloric acid is needed to decompose 50 g of limestone?

Options

A. 36.5 g
B. 73 g
C. 18.25 g
D. 91.25 g

Answer

To solve the problem of how much hydrochloric acid (HCl) is needed to decompose 50 g of limestone (calcium carbonate, CaCO3), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrochloric acid and limestone can be represented by the following balanced equation:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of CaCO3
To find the molar mass of calcium carbonate (CaCO3):
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol

Adding these together:
\[ \text{Molar mass of CaCO}_3 = 40 + 12 + 48 = 100 \text{ g/mol} \]

### Step 3: Determine the amount of HCl needed for 100 g of CaCO3
From the balanced equation, we see that 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, we need to calculate the mass of HCl required for 100 g of CaCO3.

The molar mass of HCl is:
- Hydrogen (H) = 1 g/mol
- Chlorine (Cl) = 35.5 g/mol

Thus, the molar mass of HCl is:
\[ \text{Molar mass of HCl} = 1 + 35.5 = 36.5 \text{ g/mol} \]

Since 1 mole of CaCO3 (100 g) requires 2 moles of HCl:
\[ \text{Mass of HCl required} = 2 \times 36.5 \text{ g} = 73 \text{ g} \]

### Step 4: Calculate the amount of HCl needed for 50 g of CaCO3
Since we only have 50 g of CaCO3, which is half of 100 g, we need half the amount of HCl:
\[ \text{Mass of HCl required for 50 g of CaCO}_3 = \frac{73 \text{ g}}{2} = 36.5 \text{ g} \]

### Conclusion
The mass of hydrochloric acid needed to decompose 50 g of limestone is **36.5 g**.

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