Question

Какой обьем углекислого газа выделится при сгорании 44.8 литра пропана

Answer 1

Объем пропана134.4 дм3

Answer 2

Calculating the Volume of Carbon Dioxide Produced from the Combustion of Propane

To calculate the volume of carbon dioxide (CO2) produced from the combustion of 44.8 liters of propane, we can use the balanced chemical equation for the combustion of propane and the ideal gas law.

The balanced chemical equation for the combustion of propane is: C3H8 + 5O2 → 3CO2 + 4H2O

From this equation, we can see that 1 mole of propane (C3H8) produces 3 moles of carbon dioxide (CO2) when it is completely burned.

Now, let's calculate the volume of CO2 produced using the ideal gas law, which states: PV = nRT, where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature

Given that the combustion of 1 mole of propane produces 3 moles of CO2, we can use this ratio to calculate the volume of CO2 produced from the combustion of 44.8 liters of propane.

Calculation

1. Calculate the number of moles of propane: - The molar volume of any gas at standard temperature and pressure (STP) is 22.4 liters per mole. - Therefore, the number of moles of propane = volume of propane / molar volume at STP - Number of moles of propane = 44.8 liters / 22.4 liters/mole = 2 moles

2. Calculate the number of moles of CO2 produced: - From the balanced chemical equation, 1 mole of propane produces 3 moles of CO2. - Number of moles of CO2 produced = 2 moles of propane * 3 moles of CO2/mole of propane = 6 moles

3. Calculate the volume of CO2 produced: - Using the ideal gas law, we can rearrange the formula to solve for V: V = nRT / P - Assuming standard temperature (273 K) and pressure (1 atm), and using the ideal gas constant (R = 0.0821 L·atm/(K·mol)): - V = (6 moles) * (0.0821 L·atm/(K·mol)) * (273 K) / (1 atm) - V ≈ 13.5 liters

The volume of carbon dioxide produced from the combustion of 44.8 liters of propane is approximately 13.5 liters