Calculate the mass of caco3 required to liberate 10liters of co2 at stp?

To calculate the mass of CaCO3 required to liberate 10 liters of CO2 at STP (Standard Temperature and Pressure), we need to first understand the stoichiometry of the reaction between CaCO3 and CO2.

The balanced equation for the reaction is:
CaCO3 (s) → CaO (s) + CO2 (g)

According to stoichiometry, 1 mole of CaCO3 produces 1 mole of CO2. We can use the molar volume of a gas at STP to convert between liters and moles. At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters.

Step 1: Convert 10 liters of CO2 to moles
Using the ideal gas law, moles of CO2 (n) can be calculated as:
n = V / Vm
where V is the volume of CO2 (10 liters) and Vm is the molar volume of CO2 at STP (22.4 liters/mol).

n = 10 / 22.4
n ≈ 0.4464 moles of CO2

Step 2: Convert moles of CO2 to moles of CaCO3
Since the stoichiometry of the reaction is 1:1, the moles of CaCO3 are equal to the moles of CO2.

moles of CaCO3 = 0.4464 moles

Step 3: Calculate the mass of CaCO3
To calculate the mass of CaCO3, we need to know the molar mass of CaCO3. The molar mass of CaCO3 is the sum of the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms.

Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol)
Molar mass of CaCO3 ≈ 100.09 g/mol

The mass of CaCO3 is calculated by multiplying the moles of CaCO3 by the molar mass of CaCO3:

mass of CaCO3 = moles of CaCO3 x molar mass of CaCO3
mass of CaCO3 = 0.4464 moles x 100.09 g/mol
mass of CaCO3 ≈ 44.67 grams

Therefore, approximately 44.67 grams of CaCO3 are required to liberate 10 liters of CO2 at STP.