find the volume of CO2 gas liberated at STP by thermal decomposition of 1 ton of 90% pure limestone
To find the volume of CO2 gas liberated at STP (Standard Temperature and Pressure) by thermal decomposition of limestone, we need to follow a series of steps.
1. Determine the molar mass of CO2:
The molar mass of CO2 is calculated by adding the atomic masses of one carbon (C) atom and two oxygen (O) atoms. The atomic mass of carbon is approximately 12.01 g/mol, and the atomic mass of oxygen is about 16.00 g/mol. So, the molar mass of CO2 is:
12.01 g/mol (C) + 16.00 g/mol (O) + 16.00 g/mol (O) = 44.01 g/mol
2. Convert the mass of limestone into grams:
Since we are given that we have 1 ton (1,000 kg) of limestone, we need to convert it into grams by multiplying by 1,000,000 (1 ton = 1,000 kg = 1,000,000 g).
1 ton (in grams) = 1,000,000 g
3. Determine the mass of pure limestone:
Given that the limestone is 90% pure, we need to calculate the mass of pure limestone by multiplying the total mass of limestone by 90%.
Mass of pure limestone = 90% of (mass of limestone)
4. Calculate the number of moles of pure limestone:
To calculate the mole of pure limestone, divide the mass of pure limestone by its molar mass:
Number of moles of limestone = Mass of pure limestone / Molar mass of limestone
5. Apply the stoichiometry of the reaction:
The thermal decomposition of limestone produces one mole of carbon dioxide (CO2) for every mole of limestone. Therefore, the number of moles of carbon dioxide produced will be equal to the number of moles of pure limestone.
Number of moles of CO2 = Number of moles of limestone
6. Use the Ideal Gas Law equation to find the volume:
At STP, the volume of one mole of any gas is 22.4 liters. Therefore, the volume of carbon dioxide can be calculated as:
Volume of CO2 gas at STP = Number of moles of CO2 * 22.4 liters/mole
By following these steps, you can find the volume of CO2 gas liberated at STP by thermal decomposition of 1 ton of 90% pure limestone.
CaCO3 ==> CaO + CO2
Convert 1 ton to grams. 2000 lbs x 454.6 g/lb = ?
mols CaCO3 = grams/molar mass CaCO3
Every mole CaCO3 will release 22.4 L CO2 so mols CaCO3 x 22.4 L/mol = ?