Answer:
HOPE THIS HELPS!!
Step-by-step explanation:
To calculate the percentage of pure CaCO3 in the limestone sample, we need to determine the amount of CaCO3 that reacted with the HCl.
Given:
- Mass of limestone sample = 97.51 g
- Volume of HCl solution = 16.6 cc
- Concentration of HCl solution = 0.92 N
- Sandy residue left after the reaction
First, let's calculate the number of moles of HCl used. We know that the volume of the HCl solution is 16.6 cc and the concentration is 0.92 N.
To find the moles of HCl, we can use the formula:
moles = concentration * volume
Substituting the given values, we have:
moles of HCl = 0.92 N * 0.0166 L
Next, we need to determine the number of moles of CaCO3 that reacted with the HCl. From the balanced chemical equation:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
We can see that one mole of CaCO3 reacts with 2 moles of HCl. Therefore, the moles of CaCO3 reacted would be half the moles of HCl used.
moles of CaCO3 reacted = 0.5 * moles of HCl
To calculate the moles of CaCO3 in the limestone sample, we need to convert the mass of the sample to moles. This can be done using the molar mass of CaCO3.
The molar mass of CaCO3 is:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (3 atoms)
Molar mass of CaCO3 = (40.08 + 12.01 + (16.00 * 3)) g/mol = 100.09 g/mol
moles of CaCO3 in the sample = mass of limestone sample / molar mass of CaCO3
Substituting the given value, we have:
moles of CaCO3 in the sample = 97.51 g / 100.09 g/mol
Now, to find the percentage of pure CaCO3 in the sample, we divide the moles of CaCO3 reacted by the moles of CaCO3 in the sample and multiply by 100.
percentage of pure CaCO3 = (moles of CaCO3 reacted / moles of CaCO3 in the sample) * 100
Substituting the calculated values, we have:
percentage of pure CaCO3 = (0.5 * moles of HCl) / (97.51 g / 100.09 g/mol) * 100
Calculating this expression will give us the percentage of pure CaCO3 in the sample.
Please note that the provided answer of 76.36% might not be accurate as it seems to have been rounded. It's important to perform the calculations accurately to obtain the precise result.