255 grams of octane and 1510 grams of oxygen gas are present at the beginning of reaction that goes to completion and forms carbon dioxide and water according to the following equation. 2C8H18 (l) + 2502 (g) → 16CO2 (g) + 18H2O (g) (3pts.) a) What is the limiting reactant? b) How many grams of water are formed? c) How many grams of excess reactant are consumed? d) How many grams of excess reactant are left un-reacted?
show necessary steps.
a) The limiting reactant is octane (C8H18).
b) 18 moles of water are formed, so 18 x 18.015 g/mol = 324.27 g of water are formed.
c) 1510 g of oxygen are consumed.
d) 255 g of octane are left un-reacted.
To determine the limiting reactant, we need to calculate the amount of product each reactant can produce and compare them.
a) Calculate the moles of octane (C8H18):
Molar mass of octane (C8H18) = (12.01 x 8) + (1.01 x 18) = 114.23 g/mol
Moles of octane = 255 g / 114.23 g/mol = 2.23 mol
Calculate the moles of oxygen (O2):
Molar mass of oxygen (O2) = (16 x 2) = 32.00 g/mol
Moles of oxygen = 1510 g / 32.00 g/mol = 47.19 mol
Use the balanced equation to determine the stoichiometric ratio:
2C8H18 : 25O2 : 16CO2 : 18H2O
Since the stoichiometric ratio is 2:25, we need to compare the number of moles of octane and oxygen.
Moles of oxygen needed = (2.23 mol octane) x (25 mol oxygen / 2 mol octane) = 27.88 mol
Since we have 47.19 mol of oxygen, which is greater than the moles needed, oxygen is in excess and octane is the limiting reactant.
b) Calculate the moles of water formed:
From the balanced equation, we know that for every 2 moles of octane reacted, 18 moles of water are formed.
Moles of water = (2.23 mol octane) x (18 mol water / 2 mol octane) = 20.07 mol
Calculate the mass of water:
Molar mass of water (H2O) = (1.01 x 2) + (16.00) = 18.02 g/mol
Mass of water = 20.07 mol x 18.02 g/mol = 361.42 g
Therefore, 361.42 grams of water are formed.
c) Calculate the grams of excess reactant consumed:
First, we need to calculate the moles of oxygen reacted:
Moles of oxygen reacted = moles of oxygen initially present - moles of oxygen left unreacted
Moles of oxygen reacted = 47.19 mol - 27.88 mol = 19.31 mol
Convert moles of oxygen reacted to grams:
Mass of oxygen reacted = moles of oxygen reacted x molar mass of oxygen
Mass of oxygen reacted = 19.31 mol x 32.00 g/mol = 617.12 g
Therefore, 617.12 grams of oxygen are consumed.
d) Calculate the grams of excess reactant left unreacted:
Grams of oxygen left unreacted = Mass of oxygen initially present - Mass of oxygen reacted
Grams of oxygen left unreacted = 1510 g - 617.12 g = 892.88 g
Therefore, 892.88 grams of oxygen are left unreacted.