Final answer:
The grams of oxygen needed to combust 67.9 grams of CH4, a balanced equation of the reaction is used to calculate the moles of CH4 and then convert it to moles of O2, ultimately finding that 270.72 grams of oxygen are required.
Step-by-step explanation:
The question concerns the stoichiometry of the combustion reaction of methane (CH4). To calculate the required grams of oxygen for the complete combustion of 67.9 grams of CH4, we first need to write down the balanced chemical equation:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water. Using the molar mass of methane (16.04 g/mol) and oxygen (32.00 g/mol), we can calculate the moles of methane we have:
(67.9 g CH4) / (16.04 g/mol) = 4.23 mol CH4
Since the ratio of CH4 to O2 is 1:2, we need twice the moles of oxygen:
4.23 mol CH4 × 2 = 8.46 mol O2
Now we can find the mass of the required oxygen:
(8.46 mol O2) × (32.00 g/mol) = 270.72 g O2
Therefore, 270.72 grams of oxygen are required for the complete combustion of 67.9 grams of CH4.