Calculate the number of mole of CaCl2 obtained from 25g of limestone CaCO3 in the pressure of excess HCl (C=,12, O=16, Cl=35.5)
that's "presence" not "pressure"
Figure how many moles of CaCO3 you have in 25g
Now, looking at the reaction equation:
CaCO3 + 2HCl = CaCl2 + H2O + CO2
you can see that each mole of CaCO3 produces one mole of CaCl2
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To calculate the number of moles of CaCl2 obtained from 25g of limestone CaCO3, we need to follow these steps:
1. Determine the molar mass of CaCO3:
- Calcium (Ca): 1 atom x 40 g/mol = 40 g/mol
- Carbon (C): 1 atom x 12 g/mol = 12 g/mol
- Oxygen (O): 3 atoms x 16 g/mol = 48 g/mol
- Total molar mass of CaCO3: 40 g/mol + 12 g/mol + 48 g/mol = 100 g/mol
2. Convert the mass of CaCO3 to moles:
- Number of moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
- Number of moles of CaCO3 = 25g / 100 g/mol = 0.25 mol
3. Determine the stoichiometry between CaCO3 and CaCl2:
The balanced chemical equation for the reaction between CaCO3 and HCl is:
CaCO3 + 2HCl → CO2 + CaCl2 + H2O
According to the equation, 1 mol of CaCO3 reacts to produce 1 mol of CaCl2.
4. Therefore, the number of moles of CaCl2 obtained will also be 0.25 mol.
So, from 25g of limestone CaCO3, you would obtain 0.25 mole of CaCl2 in the presence of excess HCl.