Your question's data is not correct, it should be 100 grams of limestone instead of 10 grams, You will find the correct solution below according to that.
The percentage purity of calcium carbonate is 82.5%
The balanced chemical equation is
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Because CO₂ is a gas, you can easily convert it to mole directly
No. of mol of CO₂ = volume dm³ /22.4 dm³
No. of mol of CO₂ = 8.5 dm³ / 22.4 dm³
No. of mol of CO₂ = 0.825 mol
According to equation 1 mole of CO2 is made from 1 mole of CaCO3
So, 0.825 mole of CO2 will be made from 0.825 mole of CaCO3
Mass of CaCO3 = moles of CaCO3 × Molar mass of CaCO3
Mass of CaCO3 = 0.825 × 100 grams
Mass of CaCO3 = 82.5 grams
Percentage purity of CaCO3 = (mass of CaCO3/Total mass of CaCO3) × 100
Percentage purity of CaCO3 = (82.5/100) × 100 %
Percentage purity of CaCO3 = 82.5 %
Thus we can say that as percentage purity is the percentage of mass of pure compound on per unit of total compound, the percentage purity of the calcium carbonate will be 82.5 %
Learn more about Percentage Purity here:
https://brainly.com/question/25446369
#SPJ10
