What mass of methane is burnt off completely by 224cm3 of oxygen?
To determine the mass of methane burnt off completely by a given volume of oxygen, we need to use the balanced chemical equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
From the equation, we can see that one mole of methane reacts with two moles of oxygen. Therefore, our first step is to convert the given volume of oxygen (224 cm3) to moles.
To convert from volume to moles of a gas, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
However, since we are not given the pressure or temperature, we will assume standard conditions (1 atmosphere pressure and 273.15 K). At standard conditions, one mole of any ideal gas occupies 22.4 liters (or 22,400 cm3).
So, using the conversion factor of 22,400 cm3 = 1 mole, we can calculate the number of moles of oxygen gas:
224 cm3 * (1 mole / 22,400 cm3) = 0.01 moles of oxygen
Next, we look at the balanced chemical equation and see that one mole of methane reacts with two moles of oxygen. Therefore, if 0.01 moles of oxygen are consumed, 0.01/2 = 0.005 moles of methane are burnt off.
Finally, to determine the mass of methane burnt, we need to know the molar mass of methane. The molar mass of methane (CH4) is 16 g/mol.
So, the mass of methane burnt off is:
0.005 moles * 16 g/mol = 0.08 grams of methane