Question

Сколько по объему воздуха при нормальных условиях потребуется, что бы сжечь 1 литр бензола, Плотность которого 0,88 г/(см в кубе)

Answer 1

1л=1дм3=1000см3 т.е. масса бензола 1000*0.88=880г, тогда:По уравнению реакции 2C6H6+15O2=12CO2+6H2O На два моля бензола требуется 15 молей водорода. У нас 880/78 = 11.3 моля бензола. Составляем пропорцию 2 - 15 11.3 - Х Х = 15*11,3/2 = 84,5 молей Объем кислорода равен 84.5*22,4 = 1895 литра

Answer 2

Volume of Air Required to Burn 1 Liter of Benzene at Normal Conditions

To calculate the volume of air required to burn 1 liter of benzene at normal conditions, we can use the stoichiometric equation for the combustion of benzene:

C6H6 + 15O2 → 6CO2 + 3H2O

The stoichiometric ratio of benzene to oxygen is 1:15. This means that 15 moles of oxygen are required to completely burn 1 mole of benzene.

Given that the density of benzene is 0.88 g/cm^3, we can calculate the number of moles of benzene in 1 liter using its molar mass.

The molar mass of benzene (C6H6) is approximately 78.11 g/mol.

Using the density and molar mass, we can calculate the number of moles of benzene in 1 liter: 1. Calculate the mass of 1 liter of benzene: - Mass = Volume × Density = 1 L × 0.88 g/cm^3 = 880 g 2. Calculate the number of moles of benzene: - Moles = Mass / Molar mass = 880 g / 78.11 g/mol ≈ 11.27 moles

Now, we can determine the volume of oxygen required to burn 1 liter of benzene.

Calculation of Oxygen Volume

1. Calculate the moles of oxygen required: - Moles of oxygen = Moles of benzene × Stoichiometric ratio = 11.27 moles × 15 = 169.05 moles 2. Convert moles to volume at standard conditions (0°C and 1 atm): - Volume = Moles × 22.4 L/mol (at standard conditions) = 169.05 moles × 22.4 L/mol ≈ 3786.72 L

Therefore, approximately 3786.72 liters of air at standard conditions would be required to burn 1 liter of benzene.