Certainly! Let's solve this step-by-step.
1. Identify the given data and the goal:
- Sample mass of marble: [tex]\( 0.45 \)[/tex] g
- Volume of HCl solution: [tex]\( 80 \)[/tex] ml
- Normality of HCl solution: [tex]\( 0.1 \)[/tex] N (decinormal solution)
- Find the percentage of calcium carbonate (CaCO3) in the sample.
2. Calculate the moles of HCl:
- The relationship between normality (N), volume (V) in liters, and moles (n) for the HCl solution is given by the equation:
[tex]\[
\text{moles of HCl} = \text{Normality} \times \text{Volume in liters}
\][/tex]
- Convert the volume from ml to liters:
[tex]\[
80 \, \text{ml} = \frac{80}{1000} = 0.08 \, \text{liters}
\][/tex]
- Calculate the moles of HCl:
[tex]\[
\text{moles of HCl} = 0.1 \, N \times 0.08 \, \text{liters} = 0.008 \, \text{moles}
\][/tex]
3. Determine the molar relationship between CaCO3 and HCl:
- The balanced chemical reaction between CaCO3 and HCl is:
[tex]\[
\text{CaCO3} + 2 \, \text{HCl} \rightarrow \text{CaCl2} + \text{CO2} + \text{H2O}
\][/tex]
- From the stoichiometry of the reaction, 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, to find the moles of CaCO3, we use the relationship:
[tex]\[
\text{moles of CaCO3} = \frac{\text{moles of HCl}}{2}
\][/tex]
- Calculate the moles of CaCO3:
[tex]\[
\text{moles of CaCO3} = \frac{0.008 \, \text{moles of HCl}}{2} = 0.004 \, \text{moles}
\][/tex]
4. Calculate the mass of CaCO3:
- The molar mass of CaCO3 (calcium carbonate) is approximately [tex]\( 100.0869 \)[/tex] g/mol.
- Using the moles of CaCO3, calculate the mass:
[tex]\[
\text{mass of CaCO3} = \text{moles of CaCO3} \times \text{molar mass of CaCO3}
\][/tex]
[tex]\[
\text{mass of CaCO3} = 0.004 \, \text{moles} \times 100.0869 \, \text{g/mol} = 0.4003476 \, \text{g}
\][/tex]
5. Calculate the percentage of CaCO3 in the marble sample:
- The percentage of CaCO3 is given by the ratio of the mass of CaCO3 to the mass of the sample, multiplied by 100:
[tex]\[
\text{percentage of CaCO3} = \left( \frac{\text{mass of CaCO3}}{\text{mass of sample}} \right) \times 100
\][/tex]
- Substitute the values:
[tex]\[
\text{percentage of CaCO3} = \left( \frac{0.4003476 \, \text{g}}{0.45 \, \text{g}} \right) \times 100 \approx 88.97 \%
\][/tex]
Therefore, the percentage of calcium carbonate (CaCO3) in the given sample of marble is approximately [tex]\( 88.97 \% \)[/tex].
