Calculate the number of moles volume number of molecules of carbon dioxide liberated when 200 grams of limestone decomposed .
To calculate the number of moles, volume, and number of molecules of carbon dioxide liberated when 200 grams of limestone decomposes, we need to follow these steps:
Step 1: Determine the molar mass of limestone (CaCO3)
- The molar mass of Ca = 40.08 g/mol
- The molar mass of C = 12.01 g/mol
- The molar mass of O = 16.00 g/mol
- Adding these together: 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol
Step 2: Calculate the number of moles of limestone (CaCO3)
- Use the formula: Moles = Mass / Molar Mass
- Moles = 200 g / 100.09 g/mol ≈ 1.999 moles
Step 3: Calculate the number of moles of carbon dioxide (CO2) liberated
- Limestone decomposes into one mole of CaO and one mole of CO2
- Therefore, the number of moles of CO2 liberated is also approximately 1.999 moles
Step 4: Calculate the volume of carbon dioxide (CO2) liberated
- The volume of a gas can be calculated using the ideal gas law (PV = nRT)
- Since the conditions (pressure, temperature, etc.) are not given, we cannot calculate the exact volume. However, if we assume standard conditions (1 atm pressure and 273 K temperature), we can use the molar volume of an ideal gas at standard conditions, which is 22.4 L/mol.
- Therefore, the volume of carbon dioxide liberated would be approximately 1.999 moles x 22.4 L/mol ≈ 44.76 L
Step 5: Calculate the number of molecules of carbon dioxide (CO2) liberated
- Use Avogadro's number (6.022 x 10^23 molecules/mol)
- Number of molecules = Number of moles x Avogadro's number
- Number of molecules = 1.999 moles x 6.022 x 10^23 molecules/mol ≈ 1.204 x 10^24 molecules
So, when 200 grams of limestone decompose, approximately 1.999 moles of carbon dioxide are liberated. This corresponds to a volume of approximately 44.76 liters and a number of molecules of approximately 1.204 x 10^24 molecules of carbon dioxide.