Question

1) рассчитать объем воздуха необходимый для сжигания 1.6 кг метана!! ПОМОГИТЕ

Answer 1

CH4+2O2=CO2+2H2O
v(CH4)=1600г / 16 г/моль=100 моль
v(O2)=2*100 моль=200моль
V(O2)=200 моль*22,4 л/моль= 4480 л

Answer 2

Calculating the Volume of Air Required to Burn 1.6 kg of Methane

To calculate the volume of air required to burn 1.6 kg of methane, we can use the stoichiometric ratio of methane to oxygen in the combustion reaction. The balanced chemical equation for the combustion of methane is:

CH4 + 2O2 → CO2 + 2H2O

From this equation, we can see that 1 mole of methane reacts with 2 moles of oxygen. Now, let's calculate the volume of air required.

First, we need to convert the mass of methane to moles using the molar mass of methane, which is approximately 16 g/mol.

1.6 kg = 1600 g

Now, we can calculate the moles of methane:

Moles of CH4 = Mass of CH4 / Molar mass of CH4 Moles of CH4 = 1600 g / 16 g/mol Moles of CH4 = 100 moles

According to the stoichiometry of the reaction, 1 mole of methane requires 2 moles of oxygen. Therefore, 100 moles of methane will require 200 moles of oxygen.

Next, we can calculate the volume of air required using the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.

Volume of air (in liters) = Moles of oxygen * 22.4

Let's calculate the volume of air required: Volume of air = 200 moles * 22.4 L/mol Volume of air = 4480 L

So, the volume of air required to burn 1.6 kg of methane is approximately 4480 liters.

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