Final answer:
To find the number of liters of O2 gas needed to react with 200.0 g of C8H18 at STP, convert the mass of C8H18 to moles, use stoichiometry to find the moles of O2, then convert to liters using the gas volume at STP.
Step-by-step explanation:
To find the number of liters of O2 gas needed to react with 200.0 g of C8H18 at STP, we need to convert the mass of C8H18 to moles and then use the stoichiometric ratio from the balanced equation to calculate the moles of O2. Finally, we convert the moles of O2 to liters using the fact that 1 mole of any gas at STP occupies 22.4 liters.
Step 1: Convert the mass of C8H18 to moles using its molar mass.
Molar mass of C8H18 = 8(12.01 g/mol) + 18(1.01 g/mol) = 114.23 g/mol
Moles of C8H18 = 200.0 g / 114.23 g/mol = 1.750 mol
Step 2: Use the stoichiometric ratio from the balanced equation to find the moles of O2.
From the balanced equation 2C8H18 + 25O2 -> 16CO2 + 18H2O, we see that 2 moles of C8H18 react with 25 moles of O2.
Moles of O2 = (1.750 mol C8H18)(25 mol O2 / 2 mol C8H18) = 21.875 mol O2
Step 3: Convert the moles of O2 to liters using the fact that 1 mole of any gas at STP occupies 22.4 liters.
Liters of O2 = (21.875 mol O2)(22.4 L/mol) = 489.5 L
Therefore, the correct answer is C) 33.6 liters.