calculate the mass of caco3 required to librate 10l of co2 at stp?
To calculate the mass of CaCO3 required to liberate 10L of CO2 at STP (Standard Temperature and Pressure), we need to use the concept of stoichiometry.
The balanced chemical equation for the reaction of CaCO3 to liberate CO2 is:
CaCO3 (s) → CaO (s) + CO2 (g)
From this equation, we can determine the stoichiometric relationship between CaCO3 and CO2. It tells us that for every molecule of CaCO3, one molecule of CO2 is produced.
1 mole of any gas occupies 22.4 L at STP. Therefore, 10L of CO2 is equal to 10/22.4 = 0.4464 moles of CO2.
Since the stoichiometric ratio tells us that 1 mole of CaCO3 liberates 1 mole of CO2, the moles of CaCO3 required will also be 0.4464 moles.
Now, we need to calculate the molar mass of CaCO3 (calcium carbonate) which is:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (×3 since there are three oxygen atoms)
Molar mass of CaCO3 = 40.08 + 12.01 + (16.00 * 3) = 100.09 g/mol
Finally, to find the mass of CaCO3 required, we multiply the moles of CaCO3 by its molar mass:
Mass of CaCO3 required = 0.4464 moles × 100.09 g/mol = 44.63 g
Hence, approximately 44.63 grams of CaCO3 is required to liberate 10 liters of CO2 at STP.
If your reaction is
CaCO3 -> CaO + CO2
then each mole of CaCO3 produces one mole of CO2.
So, how many moles of CO2 in 10L at STP?