Answer:
The maximum amount of CO2 that can be formed is 9.15 grams CO2
O2 is the limiting reactant
There will remain 4.82 grams of benzene
Explanation:
Step 1: Data given
Mass of benzene = 7.53 grams
Mass of oxygen gas = 8.33 grams
Molar mass of benzene = 78.11 g/mol
Molar mass oxygen gas = 32.00 g/mol
Step 2: The balanced equation
2C6H6 + 15O2 → 12CO2 + 6H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles C6H6 = 7.53 grams / 78.11 g/mol
Moles C6H6 = 0.0964 moles
Moles O2 = 8.33 grams / 32.00 g/mol
Moles O2 = 0.2603 moles
Step 4: Calculate the limiting reactant
For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles
There will remain 0.0964 - 0.0347 = 0.0617 moles benzene
This is 0.0617 moles * 78.11 g/mol = <u>4.82 grams benzene</u>
<u />
Step 5: Calculate moles CO2
For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O
For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2
Step 6: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.208 moles * 44.01 g/mol
<u>Mass CO2 = 9.15 grams</u>
