The balanced chemical equation for the combustion of methane is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
The molar mass of methane (CH4) is:
(1 x 12.01 g/mol) + (4 x 1.01 g/mol) = 16.05 g/mol
To calculate the amount of oxygen required for the combustion of 128g of methane, we need to first convert the mass of methane to moles:
128 g CH4 ÷ 16.05 g/mol = 7.98 mol CH4
According to the stoichiometry of the balanced equation, 2 moles of oxygen are required for every 1 mole of methane that reacts. Therefore, we need:
7.98 mol CH4 x (2 mol O2 / 1 mol CH4) = 15.96 mol O2
Finally, we can convert the moles of oxygen to grams:
15.96 mol O2 x 32.00 g/mol = 511.2 g O2
Therefore, 511.2 g of oxygen are required for the complete combustion of 128g of methane.