Answer:
The mass of oxygen needed to completely burn 80 grams of methane is 320 grams.
Step-by-step explanation:
The balanced chemical equation for the combustion of methane with oxygen is:
CH4 + 2O2 -> CO2 + 2H2O
From the equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to form 1 mole of carbon dioxide and 2 moles of water. We can use this information, along with the molar masses of the compounds, to calculate the mass of oxygen needed to burn 80 grams of methane.
First, we need to convert the given mqss of methane to moles. The molar mass of methane is 16.04 g/mol, so:
Moles of CH4 = Mass of CH4 / Molar mass of CH4
Moles of CH4 = 80 g / 16.04 g/mol
Moles of CH4 = 4.98 mol
Next, we can use the mole ratio from the balanced equation to calculate the moles of oxygen needed:
Moles of O2 = Moles of CH4 x (2 Moles of O2 / 1 Mole of CH4)
Moles of O2 = 4.98 mol x 2
Moles of O2 = 9.96 mol
Finally, we can convert the moles of oxygen to grqms using its molar mass of 32.00 g/mol:
Mass of O2 = Moles of O2 x Molar mass of O2
Mass of O2 = 9.96 mol x 32.00 g/mol
Mass of O2 = 319.68 g
Therefore, the mass of oxygen needed to completely burn 80 grams of methane is 319.68 grams, which we can round up to 320 grams.