30g piece of a marble was put into excess ofdil. HCl. When the reactio

1.	30g piece of a marble was put into excess ofdil. HCl. When the reaction was completed 3360 cm3 of CO2 was obtained at STP. The percentage of CaCO3 in the marble is?
Answer» Given: Mass of Marble, W = 30 gm The volume of CO2 evolved = 3360 cm³ = 3.36 L To FIND: The PERCENTAGE of CACO3 in the marble. Calculation: - At STP, 1 mole = 22.4 L - - The reaction of CaCO3 with HCl is given as: CaCO3 + HCl = CaCl2 + CO2 + H2O ⇒The mass of CaCO3 that gives 22.4 L (1 mole) of CO2 = 100 gm ⇒ The mass of CaCO3 that gives 3.36 L of CO2 = (100/22.4) × 3.36 ⇒ w = 15 gm - Percentage of CaCO3 = w/W × 100 ⇒ % CaCO3 = 15/30 × 100 ⇒ % CaCO3 = 50 % - So, the percentage of CaCO3 in the marble is 50 %.

30g piece of a marble was put into excess ofdil. HCl. When the reaction was completed 3360 cm3 of CO2 was obtained at STP. The percentage of CaCO3 in the marble is?

Answer»

Mass of Marble, W = 30 gm

The volume of CO2 evolved = 3360 cm³ = 3.36 L

The PERCENTAGE of CACO3 in the marble.

- At STP, 1 mole = 22.4 L

- The reaction of CaCO3 with HCl is given as:

CaCO3 + HCl = CaCl2 + CO2 + H2O

⇒The mass of CaCO3 that gives 22.4 L (1 mole) of CO2 = 100 gm

⇒ The mass of CaCO3 that gives 3.36 L of CO2 = (100/22.4) × 3.36

⇒ w = 15 gm

- Percentage of CaCO3 = w/W × 100

⇒ % CaCO3 = 15/30 × 100