Answer:
We can produce 6.20 grams of CO2
Explanation:
Step 1: Data given
Mass of hexane = 4.3 grams
Molar mass of hexane = 86.18 g/mol
Mass of oxygen = 7.14 grams
Molar mass of oxygen = 32.0 g/mol
Step 2: The balanced equation
2C6H14 + 19O2 → 12CO2 + 14H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles hexane = 4.3 grams / 86.18 g/mol
Moles hexane = 0.0499 moles
Moles oxygen = 7.14 grams / 32.0 g/mol
Moles oxygen = 0.2231 moles
Step 4: Calculate the limiting reactant
For 2 moles hexane we need 19 moles O2 to produce 12 moles CO2 and 14 moles H2O
Oxygen is the limiting reactant. It will completely be consumed ( 0.2231 moles). Hexane is in excess. There will react 2/19 * 0.2231 = 0.02348 moles
There will be porduced 12/19 * 0.2231 = 0.1409 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.1409 moles * 44.01 g/mol
Mass CO2 = 6.20 grams
We can produce 6.20 grams of CO2
