To find the percentage of calcium carbonate (CaCO3) in the given sample of marble, we need to go through a detailed, step-by-step procedure. Here is the solution:
### Step-by-Step Solution
1. Given Data:
- Mass of the marble sample: [tex]\(0.45 \, \text{g}\)[/tex]
- Volume of the acid (Hydrochloric Acid, HCl): [tex]\(80 \, \text{ml}\)[/tex]
- Normality of the acid: [tex]\(0.1 \, \text{N}\)[/tex] (decinormal means [tex]\(0.1 \, \text{N}\)[/tex])
2. Convert Volume to Liters:
The volume of the acid in liters is:
[tex]\[
\text{Volume of acid (L)} = 80 \, \text{ml} \times \frac{1 \, \text{L}}{1000 \, \text{ml}} = 0.08 \, \text{L}
\][/tex]
3. Calculate Moles of HCl:
Moles of HCl can be calculated using its normality and volume:
[tex]\[
\text{Moles of HCl} = \text{Normality} \times \text{Volume (in L)} = 0.1 \, \text{N} \times 0.08 \, \text{L} = 0.008 \, \text{mol}
\][/tex]
4. Stoichiometry of the Reaction:
The reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is:
[tex]\[
\text{CaCO3} + 2\text{HCl} \rightarrow \text{CaCl2} + \text{H2O} + \text{CO2}
\][/tex]
From the balanced chemical equation, 1 mole of CaCO3 reacts with 2 moles of HCl.
5. Calculate Moles of CaCO3:
Based on the stoichiometry, the moles of CaCO3 are half the moles of HCl:
[tex]\[
\text{Moles of CaCO3} = \frac{\text{Moles of HCl}}{2} = \frac{0.008 \, \text{mol}}{2} = 0.004 \, \text{mol}
\][/tex]
6. Calculate Mass of CaCO3:
The molar mass of CaCO3 (calcium carbonate) is approximately [tex]\(100 \, \text{g/mol}\)[/tex]:
[tex]\[
\text{Mass of CaCO3} = \text{Moles of CaCO3} \times \text{Molar Mass of CaCO3} = 0.004 \, \text{mol} \times 100 \, \text{g/mol} = 0.4 \, \text{g}
\][/tex]
7. Calculate Percentage of CaCO3 in the Marble Sample:
The percentage of CaCO3 in the marble sample is calculated as follows:
[tex]\[
\text{Percentage of CaCO3} = \left( \frac{\text{Mass of CaCO3}}{\text{Mass of Marble Sample}} \right) \times 100 = \left( \frac{0.4 \, \text{g}}{0.45 \, \text{g}} \right) \times 100 \approx 88.89\%
\][/tex]
### Conclusion:
The percentage of calcium carbonate (CaCO3) in the given marble sample is approximately [tex]\(88.89\%\)[/tex].
