Answer:
The maximum mass of CO2 that could be produced is 0.1925 grams
Explanation:
Step 1: Data given
Mass of methane = 0.301 grams
Mass of oxygen = 0.28 grams
Molar mass of ethane = 16.04 g/mol
Molar mass of oxygen = 32 g/mol
Step 2: The balanced equation
CH4 + 2O2 → CO2 + 2H2O
For 1 mole methane consumed, we need 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O
Step 3: Calculate moles of methane
Number of moles =Mass / Molar mass
Number of moles CH4 = 0.301 grams /16.04 g/mol
Moles CH4 = 0.0188 moles
Step 4: Calculate moles oxygen
Number of moles oxygen = 0.28 grams / 32 g/mol
Moles oxygen = 0.00875 moles
Step 5: the limiting reactant
For 1 mole methane consumed, we need 2 moles of O2
Oxygen is the limiting reactant. It will be completely consumed.
Methane is in excess. There will be consumed 0.00875 /2 = 0.004375 moles
There will remain 0.0188 - 0.004375 = 0.014425 moles
Step 6: Calculate moles of CO2
For 1 mole methane consumed, we need 2 moles of O2 to produce 1 mole of CO2
For 0.00875 moles oxygen , we have 0.004375 moles CO2
Step 7: Calculate mass of CO2
Mass CO2 = moles CO2 * Molar mass CO2
Mass CO2 =0.004375 * 44.01 g/mol = 0.1925 grams CO2
The maximum mass of CO2 that could be produced is 0.1925 grams