5.0 g of marble was added to 7.5 g dilute hydrochloric acid. After the reaction was over, it was found that 0.5 g of marble was left unused. Calculate the percentage strength of hydrochloric acid. What volume of `CO_(2)` measured at STP will be evolved in the above reaction?

Explanation

To solve the problem step by step, we need to follow these calculations:

### Step 1: Determine the mass of marble that reacted
- Initial mass of marble = 5.0 g
- Mass of marble left unused = 0.5 g
- Therefore, mass of marble that reacted = Initial mass - Unused mass
\[ \text{Mass of marble that reacted} = 5.0 \, \text{g} - 0.5 \, \text{g} = 4.5 \, \text{g} \]

### Step 2: Write the balanced chemical equation
The reaction between marble (calcium carbonate, CaCO₃) and hydrochloric acid (HCl) can be represented as:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \]

### Step 3: Calculate the moles of marble that reacted
- Molar mass of CaCO₃ = 100 g/mol
- Moles of CaCO₃ that reacted = Mass of CaCO₃ reacted / Molar mass of CaCO₃
\[ \text{Moles of CaCO}_3 = \frac{4.5 \, \text{g}}{100 \, \text{g/mol}} = 0.045 \, \text{mol} \]

### Step 4: Determine the moles of HCl that reacted
From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, moles of HCl that reacted:
\[ \text{Moles of HCl} = 2 \times \text{Moles of CaCO}_3 = 2 \times 0.045 \, \text{mol} = 0.090 \, \text{mol} \]

### Step 5: Calculate the mass of HCl that reacted
- Molar mass of HCl = 36.5 g/mol
- Mass of HCl that reacted = Moles of HCl × Molar mass of HCl
\[ \text{Mass of HCl} = 0.090 \, \text{mol} \times 36.5 \, \text{g/mol} = 3.285 \, \text{g} \]

### Step 6: Calculate the percentage strength of HCl
- Percentage strength of HCl = (Mass of HCl reacted / Total mass of HCl used) × 100
- Total mass of HCl used = 7.5 g
\[ \text{Percentage strength of HCl} = \left(\frac{3.285 \, \text{g}}{7.5 \, \text{g}}\right) \times 100 \approx 43.8\% \]

### Step 7: Calculate the volume of CO₂ evolved at STP
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, moles of CO₂ produced = moles of CaCO₃ reacted = 0.045 mol.
- Volume of CO₂ at STP = Moles of CO₂ × 22,400 cm³/mol
\[ \text{Volume of CO}_2 = 0.045 \, \text{mol} \times 22400 \, \text{cm}^3/\text{mol} = 1008 \, \text{cm}^3 \]

### Final Answers:
- Percentage strength of HCl = 43.8%
- Volume of CO₂ evolved at STP = 1008 cm³