Question

Каую масу метана можна сжечь используя кислород обьемом 112 л

Answer 1

   Х г                   112 л

   CH4       +       2O2     =     CO2       +     2H2O

n = 1 моль          n = 2 моль

Mr = 16 г/моль   Vm = 22,4 л/моль

m = 16 г               V = 44,8 л

Х г СН4     -    112 л О2

16 г СН4   -    44,8 л О2

m(CH4) = 112 * 16 / 44,8 = 40 г

Answer 2

Methane Combustion with Oxygen

To calculate the amount of methane that can be burned using a given volume of oxygen, we can use the balanced chemical equation for the combustion of methane:

CH4 + 2O2 → CO2 + 2H2O

According to the stoichiometry of this reaction, 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

Given that the volume of the oxygen is 112 L, we can use the ideal gas law to convert the volume of oxygen to moles. The ideal gas law is given by the equation:

PV = nRT

Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature

Assuming standard temperature and pressure (STP), the molar volume of an ideal gas is 22.4 L/mol. Therefore, the number of moles of oxygen can be calculated as:

n = V / 22.4

Substituting the given volume of oxygen: n = 112 L / 22.4 L/mol = 5 moles

Now, according to the balanced chemical equation, 1 mole of methane reacts with 2 moles of oxygen. Therefore, the amount of methane that can be burned can be calculated as:

Moles of methane = 5 moles / 2 = 2.5 moles

So, the amount of methane that can be burned using 112 L of oxygen is approximately 2.5 moles.