Question

Розрахуйте який об'єм кисню необхідний для спалювання 1 л пропану, 1 л бутану, 1 л пентану. Зробити висновок про споживання кисню під час спалювання різних алканів. Об'єми всіх газів виміряно за однакових умов.

Answer 1

Ответ:

Объяснение:

Реакция горения пропана:

C3H8+5O2=3CO2+4H2O

V(O2)=V(C3H8)*5=5 л

Реакция горения бутана:

C4H10+6.5O2=4CO2+5H2O

V(O2)=V(C4H10)*6.5=6.5 л

Реакция горения пентана:

C5H12+8O2=5CO2+6H2O

V(O2)=V(C5H12)*8=8 л

Из этих реакций видно, что объем кислорода, необходимый для полного сгорания алкана, увеличивается с ростом предельного углеводорода. Рассмотрим общее уравнение горения алканов:

CnH(2n+2)+(3n+1)/2*O2=nCO2+(n+1)H2O

Из записанного уравнения видно, что при росте n (т.е. росте углеводородной цепочки), увеличивается объем необходимого кислорода.

Answer 2

Calculating Oxygen Volume for Combustion of Propane, Butane, and Pentane

To calculate the volume of oxygen needed for the combustion of 1 liter of propane, 1 liter of butane, and 1 liter of pentane, we can use the balanced chemical equations for their combustion reactions. The general chemical equation for the combustion of alkanes is:

Alkane + Oxygen → Carbon Dioxide + Water

1. Propane (C3H8): - The balanced chemical equation for the combustion of propane is: ``` C3H8 + 5O2 → 3CO2 + 4H2O ``` - According to the balanced equation, 5 moles of oxygen are required to burn 1 mole of propane.

2. Butane (C4H10): - The balanced chemical equation for the combustion of butane is: ``` C4H10 + 6.5O2 → 4CO2 + 5H2O ``` - According to the balanced equation, 6.5 moles of oxygen are required to burn 1 mole of butane.

3. Pentane (C5H12): - The balanced chemical equation for the combustion of pentane is: ``` C5H12 + 8O2 → 5CO2 + 6H2O ``` - According to the balanced equation, 8 moles of oxygen are required to burn 1 mole of pentane.

Conclusion on Oxygen Consumption during Combustion of Different Alkanes

From the balanced chemical equations, we can conclude that: - Propane requires 5 moles of oxygen for combustion. - Butane requires 6.5 moles of oxygen for combustion. - Pentane requires 8 moles of oxygen for combustion.

This indicates that as the number of carbon atoms in the alkane molecule increases, the amount of oxygen required for combustion also increases. This is due to the higher carbon-to-hydrogen ratio in longer alkanes, leading to the need for more oxygen to completely oxidize the carbon and hydrogen atoms in the molecules.

Therefore, the volume of oxygen needed for the combustion of 1 liter of propane, 1 liter of butane, and 1 liter of pentane can be calculated based on the molar volume of gases at standard temperature and pressure (STP), which is approximately 22.4 liters per mole.

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