Question

Какой объём углекислого газа образуется при прокаливании 150 г карбоната бария, содержащего 10% примесей

Answer 1

BaCO3=BaO+CO2
в 100 г чистого 90
в 150 ------ x
m(BaCO3)=$\frac{90*150}{100}$ =135 г
M(BaCO3)=137+12+16*3=197
n=m/M=135/197=0.7 моль
V(CO2)=n*Vm=0.7 моль*22.4 л=15,68 л

Answer 2

Calculation of Carbon Dioxide Gas Produced from Barium Carbonate

To calculate the volume of carbon dioxide gas produced when 150g of barium carbonate containing 10% impurities is heated, we can use the following steps:

1. Calculate the moles of pure barium carbonate: - The molar mass of barium carbonate (BaCO3) is 197.34 g/mol. - Calculate the moles of pure barium carbonate using the formula: ``` moles = mass / molar mass ```

2. Determine the moles of carbon dioxide produced: - The balanced chemical equation for the decomposition of barium carbonate is: ``` BaCO3 -> BaO + CO2 ``` - From the equation, it is clear that 1 mole of barium carbonate produces 1 mole of carbon dioxide.

3. Calculate the volume of carbon dioxide gas at standard temperature and pressure (STP): - At STP (0°C and 1 atm), 1 mole of any gas occupies 22.4 liters.

Let's perform the calculations based on the given information.

Calculation Steps

1. Calculate the moles of pure barium carbonate: - Molar mass of barium carbonate (BaCO3) = 197.34 g/mol - Moles of pure barium carbonate = 150g / 197.34 g/mol - Moles of pure barium carbonate = 0.760 moles

2. Determine the moles of carbon dioxide produced: - Since 1 mole of barium carbonate produces 1 mole of carbon dioxide, the moles of carbon dioxide produced will also be 0.760 moles.

3. Calculate the volume of carbon dioxide gas at STP: - Volume of carbon dioxide gas at STP = 0.760 moles * 22.4 L/mole - Volume of carbon dioxide gas at STP = 17.024 liters

Conclusion

When 150g of barium carbonate, containing 10% impurities, is heated, approximately 17.024 liters of carbon dioxide gas is produced at standard temperature and pressure (STP).