Question

Определите объём воздуха необходимый для сгорания 10 моль бутана

Answer 1

Дано:
n(C4H10) = 10 моль
Найти:
V(возд)-?

φ(О2) = 0,21
Vm  = 22,4 л/моль

2С4Н10 + 13О2 = 8СО2 + 10Н2О
Из УХР следует, что n(O2) = 6.5n(C4H10)
n(O2) = 6,5*10 моль = 65 моль
V(O2) = n*Vm = 65 моль*22,4 л = 1456 л
V(возд) = V(O2)/φ(O2) = 1456 л/0,21 = 6933 л ≈7 м³
Ответ: 7 м³

Answer 2

Calculating the Volume of Air Needed to Burn 10 Moles of Butane

To calculate the volume of air needed to burn 10 moles of butane, we can use the balanced chemical equation for the combustion of butane. The balanced chemical equation for the combustion of butane is as follows:

C4H10 + 13O2 → 8CO2 + 10H2O

From the balanced chemical equation, we can see that 1 mole of butane reacts with 13 moles of oxygen. Therefore, to calculate the volume of air (oxygen) needed to burn 10 moles of butane, we can use the stoichiometry of the reaction.

Using the ideal gas law, we can relate the number of moles of a gas to its volume at standard temperature and pressure (STP). At STP, 1 mole of any gas occupies 22.4 liters.

So, the volume of air needed to burn 10 moles of butane can be calculated as follows:

Volume of air (O2) = 10 moles of butane * (13 moles of O2 / 1 mole of butane) * 22.4 liters/mole

Volume of air (O2) = 10 * 13 * 22.4 liters

Volume of air (O2) = 3640 liters

Therefore, the volume of air needed to burn 10 moles of butane is 3640 liters.

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