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A 1.502 g dry sample containing only CaCO3 and CaCl2 was dissolved in 25.00 ml of 0.925 m HCI solution. What is the CaCl2 percentage in the sample, if 22 ml of 0.09312 m NaOH was used to titrate the excess HCI? A) 30% b) 70% c) 40% d) 60%
Accepted Answer
Here's how to solve the problem:
1. Calculate moles of NaOH used:
- Moles NaOH = (Volume NaOH)
(Molarity NaOH) = (22 mL)
(0.09312 mol/L) = 0.0020486 mol
2. Calculate moles of excess HCl:
- Moles NaOH = Moles excess HCl (since they react 1:1)
- Therefore, moles excess HCl = 0.0020486 mol
3. Calculate moles of HCl initially present:
- Moles HCl = (Volume HCl)
(Molarity HCl) = (25.00 mL)
(0.925 mol/L) = 0.023125 mol
4. Calculate moles of HCl reacted with CaCO3:
- Moles HCl reacted = Moles initial HCl - Moles excess HCl = 0.023125 mol - 0.0020486 mol = 0.0210764 mol
5. Calculate moles of CaCO3:
- CaCO3 + 2HCl → CaCl2 + CO2 + H2O
- The reaction shows that 1 mole of CaCO3 reacts with 2 moles of HCl.
- Moles CaCO3 = (Moles HCl reacted) / 2 = 0.0210764 mol / 2 = 0.0105382 mol
6. Calculate mass of CaCO3:
- Mass CaCO3 = (Moles CaCO3)
(Molar mass CaCO3) = (0.0105382 mol)
(100.09 g/mol) = 1.054 g
7. Calculate mass of CaCl2:
- Mass CaCl2 = Mass sample - Mass CaCO3 = 1.502 g - 1.054 g = 0.448 g
8. Calculate the percentage of CaCl2 in the sample:
- Percentage CaCl2 = (Mass CaCl2 / Mass sample)
100% = (0.448 g / 1.502 g)
100% = 29.83%
Therefore, the closest answer is A) 30%
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