What is the volume of oxygen required to burn completely 45cm3 of methane.
To calculate the volume of oxygen required to burn completely 45 cm3 of methane, we need to balance the chemical equation for the combustion of methane (CH4 + O2 -> CO2 + H2O).
The equation shows that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.
From the balanced equation, we can see that the stoichiometric ratio between methane and oxygen is 1:2. This means that for every one volume of methane, we need two volumes of oxygen.
Therefore, since we have 45 cm3 of methane, we would need (2 x 45 cm3) = 90 cm3 of oxygen to burn completely 45 cm3 of methane.