To determine the volume of CO2 gas produced when 20 g of 20% pure CaCO3 is completely heated, we can follow these steps:
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Calculate the mass of CaCO3 in the sample: Since the purity of CaCO3 is 20%, we can find the amount of pure CaCO3 in 20 g of the sample: \[ \text{Mass of CaCO3} = 20 , \text{g} \times 0.20 = 4 , \text{g} \]
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Determine the molar mass of CaCO3: The molar mass of calcium carbonate (CaCO3) can be calculated as follows:
- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O3: 3 × 16.00 g/mol = 48.00 g/mol
Therefore, the total molar mass of CaCO3 is: \[ 40.08 + 12.01 + 48.00 = 100.09 , \text{g/mol} \]
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Calculate the number of moles of CaCO3: Now we can find the number of moles of CaCO3 in 4 g: \[ \text{Number of moles of CaCO3} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 , \text{g}}{100.09 , \text{g/mol}} \approx 0.03996 , \text{mol} \]
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Chemical reaction for thermal decomposition: When CaCO3 is heated, it decomposes according to the following reaction: \[ \text{CaCO3} , (s) \rightarrow \text{CaO} , (s) + \text{CO2} , (g) \] From this reaction, we see that 1 mole of CaCO3 produces 1 mole of CO2.
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Calculate the moles of CO2 produced: Since 1 mole of CaCO3 produces 1 mole of CO2, the moles of CO2 produced will be the same as the moles of CaCO3, which is approximately 0.03996 mol.
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Calculate the volume of CO2 at standard temperature and pressure (STP): At STP (0°C and 1 atm), 1 mole of an ideal gas occupies approximately 22.4 liters. \[ \text{Volume of CO2} = \text{Number of moles of CO2} \times \text{Volume at STP} = 0.03996 , \text{mol} \times 22.4 , \text{L/mol} \approx 0.894 , \text{L} \]
So, the volume of CO2 gas produced when 20 g of 20% pure CaCO3 is completely heated is approximately 0.894 liters.