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?
a) Octane is the limiting reactant.
b) 18 x 18 = 324 grams of water are formed.
c) 255 grams of octane are consumed.
d) 1510 grams of oxygen gas are left un-reacted.
To determine the limiting reactant, you need to compare the number of moles of each reactant and see which one has the lowest mole ratio to the products. Here's how to calculate it:
Step 1: Calculate the moles of octane (C8H18) and oxygen gas (O2):
- Moles of octane = mass of octane / molar mass of octane
- Moles of oxygen gas = mass of oxygen gas / molar mass of oxygen gas
The molar mass (in grams/mol) of octane (C8H18) can be calculated by adding up the molar masses of carbon (C) and hydrogen (H). Similarly, the molar mass of oxygen gas (O2) is just the molar mass of oxygen (O) multiplied by 2.
Step 2: Calculate the mole ratio between the reactants and products:
From the balanced equation, you can see that the mole ratio between octane and water is 2:18, and the mole ratio between oxygen gas and water is 2502:18.
Step 3: Compare the mole ratios:
Divide the moles of each reactant by its respective mole ratio to water. The reactant with the smaller resulting value is the limiting reactant.
Step 4: Calculate the grams of water formed:
Multiply the moles of the limiting reactant by the molar mass of water to find the grams of water formed.
Step 5: Calculate the grams of excess reactant consumed:
Subtract the moles of the limiting reactant from the initial moles of each reactant and convert it to grams using the respective molar masses.
Step 6: Calculate the grams of excess reactant left unreacted:
Multiply the moles of the excess reactant by its molar mass to find the grams of the excess reactant left unreacted.
Let's go through each step to find the answers:
a) To determine the limiting reactant:
- Moles of octane = 255 g / molar mass of octane
- Moles of octane = 255 g / (molar mass of C8H18)
- Moles of oxygen = 1510 g / molar mass of oxygen
- Moles of oxygen = 1510 g / (molar mass of O2)
b) To calculate the grams of water formed:
- Determine the moles of the limiting reactant (whichever one had the smallest value when comparing moles/ratios).
- Multiply the moles of the limiting reactant by the molar mass of water.
- Grams of water formed = moles of limiting reactant x molar mass of water
c) To calculate the grams of excess reactant consumed:
- Subtract the moles of the limiting reactant from the initial moles of each reactant.
- Convert the resulting moles to grams by multiplying by their respective molar masses.
d) To calculate the grams of excess reactant left unreacted:
- Multiply the moles of the excess reactant (the one that is not the limiting reactant) by its molar mass.
Using these steps, you should be able to find the answers to parts a, b, c, and d of the given question.
To determine the limiting reactant, we need to calculate the moles of each reactant and compare them to their stoichiometric coefficients in the balanced equation.
a) First, we calculate the moles of octane (C8H18) and oxygen (O2) using their respective molar masses:
Molar mass of octane (C8H18) = (12.01 g/mol * 8) + (1.01 g/mol * 18) = 114.23 g/mol
Moles of octane = grams of octane / molar mass of octane = 255 g / 114.23 g/mol
Molar mass of oxygen (O2) = (16.00 g/mol * 2) = 32.00 g/mol
Moles of oxygen = grams of oxygen / molar mass of oxygen = 1510 g / 32.00 g/mol
b) The balanced equation tells us that 2 moles of octane produce 18 moles of water. Therefore, we can use stoichiometry to find the moles of water produced:
Moles of water produced = (moles of octane) / 2 * 18
c) To find the moles of excess reactant consumed, we need to compare the moles of oxygen consumed based on the balanced equation:
Moles of oxygen consumed = (moles of octane) / 2 * 2502
d) Finally, to find the grams of excess reactant left unreacted, we can calculate the mass of the excess reactant initially present minus the mass of the reactant consumed:
Mass of excess reactant left un-reacted = initial mass of excess reactant - mass of excess reactant consumed
Note: To proceed with b, c, and d, we need to know which of the two reactants is the limiting reactant.
To determine the limiting reactant, we can compare the moles of each reactant. The reactant with fewer moles is the limiting reactant.