Question

Какой объем оксида углерода (IV) образуется при взаимодействии 60 г карбоната кальция, содержащего 10% примесей, с соляной кислотой?

Answer 1

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Answer 2

Calculating the Volume of Carbon Dioxide (IV) Formed

To calculate the volume of carbon dioxide (IV) formed when 60g of calcium carbonate, containing 10% impurities, reacts with hydrochloric acid, we can use the following steps:

1. Calculate the moles of calcium carbonate (CaCO3): - The molar mass of CaCO3 is 100.09 g/mol. - Therefore, the moles of CaCO3 can be calculated as: ``` moles = mass / molar mass ```

2. Determine the limiting reactant: - The reaction between calcium carbonate and hydrochloric acid can be represented as: ``` CaCO3 + 2HCl -> CaCl2 + H2O + CO2 ``` - The stoichiometry of the reaction shows that 1 mole of CaCO3 reacts with 2 moles of HCl. - Calculate the moles of HCl that can react with the moles of CaCO3.

3. Calculate the volume of carbon dioxide (IV) formed: - Use the ideal gas law, PV = nRT, to calculate the volume of CO2 produced.

Let's proceed with the calculations.

Calculation Steps

1. Calculate the moles of calcium carbonate (CaCO3): - Molar mass of CaCO3 = 100.09 g/mol - Moles of CaCO3 = 60g / 100.09 g/mol = 0.599 moles.

2. Determine the limiting reactant: - Since the impurities are not specified, we'll assume that the 60g of calcium carbonate contains 10% impurities, meaning the actual mass of pure CaCO3 is 54g. - Moles of CaCO3 = 54g / 100.09 g/mol = 0.539 moles. - The stoichiometry of the reaction shows that 1 mole of CaCO3 reacts with 2 moles of HCl. - Moles of HCl required = 2 * 0.539 = 1.078 moles.

3. Calculate the volume of carbon dioxide (IV) formed: - Assuming standard temperature and pressure (STP), which is 0°C and 1 atm, the molar volume of a gas is 22.4 L/mol at STP. - Volume of CO2 formed = 1.078 moles * 22.4 L/mol = 24.15 L.

Conclusion

When 60g of calcium carbonate, containing 10% impurities, reacts with hydrochloric acid, the volume of carbon dioxide (IV) formed at standard temperature and pressure (STP) is approximately 24.15 liters.