Question

Сколько литров углекислого газа получится при разложении 250 г карбоната кальция, если объемная доля выхода продукта реакции составляет 90% от теоретически возможного?

Answer 1

m(CaCO3) = 250 g

W(выхода CO2) = 90%

V(CO2) - ?

n(CaCO3) = 250 g / 100 g / mol = 2.5 mol

CaCO3 = CaO + CO2

n(CaCO3) = n(CO2) = 2.5 mol

V(CO2 теоретического) = 2.5 mol * 22.4 l / mol = 56 l

V(CO2 практического) = 56 l * 0.9 = 50.4 l
Ответ: 50.4 l .

Answer 2

Calculating the Volume of Carbon Dioxide Produced from the Decomposition of Calcium Carbonate

To calculate the volume of carbon dioxide produced from the decomposition of 250 g of calcium carbonate, we can use the following steps:

1. Determine the Molar Mass of Calcium Carbonate: - The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol.

2. Calculate the Number of Moles of Calcium Carbonate: - Number of moles = Mass / Molar mass - Number of moles = 250 g / 100.09 g/mol - Number of moles ≈ 2.499 moles

3. Determine the Stoichiometry of the Reaction: - The balanced chemical equation for the decomposition of calcium carbonate is: `CaCO3 → CaO + CO2` - From the balanced equation, we can see that 1 mole of calcium carbonate produces 1 mole of carbon dioxide.

4. Calculate the Theoretical Volume of Carbon Dioxide: - The volume of 1 mole of any gas at standard temperature and pressure (STP) is 22.4 liters. - The theoretical volume of carbon dioxide = Number of moles × 22.4 L/mol - The theoretical volume of carbon dioxide ≈ 2.499 moles × 22.4 L/mol ≈ 56.18 liters

5. Adjust for the Yield: - The given information states that the actual yield is 90% of the theoretical yield.

6. Calculate the Actual Volume of Carbon Dioxide: - Actual volume = Theoretical volume × Yield percentage - Actual volume ≈ 56.18 liters × 90% ≈ 50.562 liters

Therefore, the volume of carbon dioxide produced from the decomposition of 250 g of calcium carbonate, with a yield of 90%, is approximately 50.562 liters.