Question

Какой объем углекислого газа образуется если 56 л пропана сожгли в кислороде?

Answer 1

Решение:
C3H8+5O2=3CO2+4H2O
n(C3H8)=56/22.4=2.5моль
n(CO2)=2.5/1 : x/3 =7.5моль
V(CO2)=n*Vm=7.5*22.4=168л

Answer 2

C3h8+5O2->3Co2+4h20
Объем co2 в три раза больше объема пропана , следовательно 56*3=168 литров
Ответ:168 л

Answer 3

Calculating the Volume of Carbon Dioxide Produced

To calculate the volume of carbon dioxide produced when 56 L of propane is burned in oxygen, we can use the balanced chemical equation for the combustion of propane:

C3H8 + 5O2 → 3CO2 + 4H2O

From the balanced equation, we can see that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2).

Calculation Steps

1. Calculate the moles of propane (C3H8) using the given volume and molar volume at standard temperature and pressure (STP). 2. Use the mole ratio from the balanced chemical equation to find the moles of carbon dioxide (CO2) produced. 3. Convert the moles of carbon dioxide to volume using the molar volume at STP.

Calculation

1. Moles of Propane (C3H8) - Given: Volume of propane = 56 L - Molar volume at STP (standard temperature and pressure) = 22.4 L/mol - Moles of propane = Volume of propane / Molar volume at STP - Moles of propane = 56 L / 22.4 L/mol - Moles of propane = 2.5 moles

2. Moles of Carbon Dioxide (CO2) - Using the mole ratio from the balanced chemical equation: - 1 mole of propane produces 3 moles of carbon dioxide - Moles of carbon dioxide = Moles of propane * (3 moles CO2 / 1 mole C3H8) - Moles of carbon dioxide = 2.5 moles * 3 - Moles of carbon dioxide = 7.5 moles

3. Volume of Carbon Dioxide (CO2) - Molar volume at STP = 22.4 L/mol - Volume of carbon dioxide = Moles of carbon dioxide * Molar volume at STP - Volume of carbon dioxide = 7.5 moles * 22.4 L/mol - Volume of carbon dioxide = 168 L

Therefore, the volume of carbon dioxide produced when 56 L of propane is burned in oxygen is 168 liters.