Question

Объем углекислого газа при нормальных условиях который выделится при разложение 300 г извесника содержащего 15% примеси

Answer 1

В приложении решение к задаче

Answer 2

Calculating the Volume of Carbon Dioxide Released

To calculate the volume of carbon dioxide (CO2) released when 300g of limestone containing 15% impurities decomposes under normal conditions, we can use the following chemical equation for the decomposition of limestone:

CaCO3(s) → CaO(s) + CO2(g)

The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CO2 is 44.01 g/mol. Given that the limestone contains 15% impurities, we can calculate the mass of pure CaCO3 in 300g of limestone.

Calculation Steps

1. Calculate the mass of pure CaCO3 in 300g of limestone: - Mass of CaCO3 = 300g * (1 - 0.15) = 300g * 0.85 = 255g

2. Convert the mass of pure CaCO3 to moles: - Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3 = 255g / 100.09 g/mol = 2.549 moles

3. Use the stoichiometry of the reaction to find the moles of CO2 produced: - According to the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CO2.

4. Calculate the volume of CO2 at standard temperature and pressure (STP): - 1 mole of any gas occupies 22.4 liters at STP.

Conclusion

The volume of carbon dioxide released when 300g of limestone containing 15% impurities decomposes under normal conditions is approximately 57.1 liters.

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