resulting solution acidic or bask 4.94 A 1.248 g sample of limestone rock is pulverized and thin treated with 30.00 mL of 1.035 M HCl solution. The exce s acid then requires 11 .56 mL of 1 .010 M NaOH for neutraliz.- tion. Calculate the percentage by mass of calcium carbonate in the rock , assuming that it is the only substance reacting with the HCl solution.

To calculate the percentage by mass of calcium carbonate (CaCO₃) in the limestone rock, we need to follow these steps:<br /><br />1. **Determine the molesCl used:**<br /> \[<br /> \text{Volume of HCl solution} = 30.00 \, \text{mL} = 0.03000 \, \text{L}<br /> \]<br /> \[<br /> \text{Concentration of HCl solution} = 1.035 \, \text{M}<br /> \]<br />[<br /> \text{Moles of HCl} = \text{Volume} \times \text{Concentration} = 0.03000 \, \text{L} \times 1.035 \, \text{M} = 0.03105 \, \text{moles}<br /> \]<br /><br />2. **Determine the moles of NaOH used for neutralization:**<br /> \[<br /> \text{Volume of NaOH solution} = 11.56 \, \text{mL} = 0.01156 \, \text{L}<br /> \]<br /> \[<br /> \text{Concentration of NaOH solution} = 1.010 \, \text{M}<br /> \]<br /> \[<br /> \text{Moles of NaOH} = \text{Volume} \times \text{Concentration} = 0.01156 \, \text{L} \times 1.010 \, \text{M} = 0.01167 \, \text{moles}<br /> \]<br /><br />3. **Determine the moles of HCl remaining after reacting with CaCO₃:**<br /> Since the moles of NaOH used for neutralization are equal to the moles of HCl remaining after reacting with CaCO₃:<br /> \[<br /> \text{Moles of HCl remaining} = 0.01167 \, \text{moles}<br /> \]<br /><br />4. **Determine the moles of HCl that reacted with CaCO₃:**<br /> \[<br /> \text{Moles of HCl that reacted with CaCO₃} = \text{Total moles of HCl} - \text{Moles of HCl remaining} = 0.03105 \,{moles} - 0.01167 \, \text{moles} = 0.01938 \, \text{moles}<br /> \]<br /><br />5. **Determine the moles of CaCO₃ in the rock:**<br /> The reaction between HCl and CaCO₃ is a 1:1 molar ratio:<br /> \[<br /> \text{CaCO₃} + 2 \text{HCl} \rightarrow \text{CaCl₂} + \text{H₂O} + \text{CO₂}<br /> \]<br /> Therefore, the moles of CaCO₃ are equal to the moles of HCl that reacted:<br /> \[<br /> \text{Moles of CaCO₃} = 0.01938 \, \text{moles}<br /> \]<br /><br />6. **Determine the mass of CaCO₃ in the rock:**<br /> \[<br /> \text{Molar mass of CaCO₃} = 100.09 \, \text{g/mol}<br /> \]<br /> \[<br /> \text{Mass of CaCO₃} = \text{Moles of CaCO₃} \text{Molar mass of CaCO₃} = 0.01938 \, \text{moles} \times 100.09 \, \text{g/mol} = 1.940 \, \text{g}<br /> \]<br /><br />7. **Determine the percentage by mass of CaCO₃ in the rock:**<br /> \[<br /> \text{Mass of limestone rock sample} = 1.248 \, \text{g}<br /> \]<br /> \[<br /> \text{Percentage by mass of CaCO₃} = \left( \frac{\text{Mass of CaCO₃}}{\text{Mass of limestone rock sample}} \right) \times 100\% = \left( \frac{1.940 \, \text{g}}{1.248 \, \text{g}} \right) \times 100\% = 155.56\%<br /> \]<br /><br />Since the percentage by mass cannot be greater than 100%, there might be an error in the calculation or assumptions. Please recheck the steps and data provided.