4g of a mixture of CaCO3 and sand is treated with an excess of HCl and 0.88g of CO2 is produced. What is the percentage of CaCO3 in the mixture

44gms of CO2

is produced by (40+12+48)100gms of CaCO3
0.88gm of CO2
must be produced by (100/44)×0.88= 2gm of CaCO3
So the amount of pure CaCO3
present in 4gm of impure sample is 2gms.
So percentage of purity = (2/4) ×100= 50

it is right

It's right

To find the percentage of CaCO3 in the mixture, you need to use the stoichiometry of the reaction between CaCO3 and HCl. Here's how you can calculate it step by step:

1. Calculate the molar mass of CO2:
The molar mass of CO2 is 12.01 g/mol (for carbon) + 2 * 16.00 g/mol (for oxygen) = 44.01 g/mol.

2. Convert the mass of CO2 produced to moles:
Divide the mass of CO2 produced (0.88 g) by the molar mass of CO2 to get the number of moles of CO2 produced:
Moles of CO2 = 0.88 g / 44.01 g/mol ≈ 0.02 mol.

3. Use the stoichiometry of the reaction to determine the moles of CaCO3:
From the balanced equation of the reaction between CaCO3 and HCl, we know that:
1 mol of CaCO3 reacts with 1 mol of CO2.
So, the moles of CaCO3 is equal to the moles of CO2 produced:
Moles of CaCO3 = 0.02 mol.

4. Calculate the molar mass of CaCO3:
The molar mass of CaCO3 is 40.08 g/mol (for calcium) + 12.01 g/mol (for carbon) + 3 * 16.00 g/mol (for oxygen) = 100.09 g/mol.

5. Calculate the mass of CaCO3 in the mixture:
Multiply the moles of CaCO3 by the molar mass of CaCO3 to get the mass of CaCO3 in the mixture:
Mass of CaCO3 = 0.02 mol * 100.09 g/mol ≈ 2.00 g.

6. Calculate the percentage of CaCO3 in the mixture:
Divide the mass of CaCO3 by the initial mass of the mixture and multiply by 100 to get the percentage:
Percentage of CaCO3 = (2.00 g / 4.00 g) * 100 ≈ 50%.

Therefore, the percentage of CaCO3 in the mixture is approximately 50%.

Convert 0.88 g CO2 to CaCO3.

That's 0.88 x (molar mass CaCO3/molar mass CO2) = ?g CaCO3
Then % CaCO3 in sample = (?g CaCO3/mass sample)*100 = ?