Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 5.71 g of octane is mixed with 5.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Please show steps I am very confused on how to work this!
start with a balanced reaction.
Convert the 5.71g octane, and 5g of O2 to moles. Which is the limiting reactant? That reactant will determine the products.
I wlll be happy to check your work.
To calculate the maximum mass of carbon dioxide produced in the reaction between octane and oxygen, we need to use the concept of stoichiometry, which relates the amounts of reactants and products in a chemical reaction.
Here are the steps to solve this problem:
Step 1: Write the balanced chemical equation for the reaction:
2 C8H18 (l) + 25 O2 (g) -> 16 CO2 (g) + 18 H2O (g)
Step 2: Determine the molar masses of the relevant compounds:
- Octane (C8H18): 114.22 g/mol
- Oxygen (O2): 32.00 g/mol
- Carbon dioxide (CO2): 44.01 g/mol
Step 3: Calculate the number of moles of each reactant:
- Moles of octane:
5.71 g octane * (1 mol octane / 114.22 g octane) = 0.0500 mol octane (rounded to 4 significant digits)
- Moles of oxygen:
5.00 g oxygen * (1 mol oxygen / 32.00 g oxygen) = 0.156 mol oxygen (rounded to 3 significant digits)
Step 4: Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the mole ratios of octane and oxygen from the balanced equation.
From the balanced equation, we see that 2 moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide. So the mole ratio of octane to oxygen is 2:25.
Using this ratio, we can calculate the number of moles of oxygen required to react with the given 0.0500 moles of octane:
0.0500 mol octane * (25 mol oxygen / 2 mol octane) = 0.3125 mol oxygen
Since we only have 0.156 mol of oxygen, which is less than the required amount, oxygen is the limiting reactant.
Step 5: Calculate the maximum mass of carbon dioxide produced:
Now that we know that oxygen is the limiting reactant, we can determine the maximum amount of carbon dioxide produced using the mole ratio between oxygen and carbon dioxide from the balanced equation.
From the balanced equation, we see that 25 moles of oxygen react to produce 16 moles of carbon dioxide.
Using this ratio, we can calculate the number of moles of carbon dioxide produced:
0.156 mol oxygen * (16 mol carbon dioxide / 25 mol oxygen) = 0.09984 mol carbon dioxide
Finally, we can determine the maximum mass of carbon dioxide produced by multiplying the number of moles by the molar mass of carbon dioxide:
0.09984 mol carbon dioxide * 44.01 g/mol = 4.39 g carbon dioxide (rounded to 3 significant digits)
Therefore, the maximum mass of carbon dioxide that could be produced by the chemical reaction is 4.39 g.
To calculate the maximum mass of carbon dioxide produced, we need to determine which reactant (octane or oxygen) is the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, thereby limiting the amount of product that can be formed.
We'll start by calculating the number of moles of each reactant using their respective molar masses.
The molar mass of octane (C8H18) is:
(8 * atomic mass of carbon) + (18 * atomic mass of hydrogen) = (8 * 12.01 g/mol) + (18 * 1.01 g/mol) = 114.23 g/mol
The number of moles of octane is given by:
moles of octane = mass of octane / molar mass of octane = 5.71 g / 114.23 g/mol = 0.05 mol
The molar mass of oxygen (O2) is:
(2 * atomic mass of oxygen) = (2 * 16.00 g/mol) = 32.00 g/mol
The number of moles of oxygen is given by:
moles of oxygen = mass of oxygen / molar mass of oxygen = 5.0 g / 32.00 g/mol = 0.16 mol
Next, we need to determine the stoichiometric ratio between octane and carbon dioxide. From the balanced chemical equation:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
We can see that 2 moles of octane produce 16 moles of carbon dioxide.
Using the stoichiometric ratio, we can calculate the maximum number of moles of carbon dioxide that could be produced:
moles of carbon dioxide = (moles of octane) * (moles of carbon dioxide / moles of octane)
= 0.05 mol * (16 mol CO2 / 2 mol C8H18) = 0.4 mol CO2
Finally, we can calculate the maximum mass of carbon dioxide produced by multiplying the number of moles calculated above by the molar mass of carbon dioxide:
mass of carbon dioxide = moles of carbon dioxide * molar mass of CO2
= 0.4 mol * 44.01 g/mol = 17.6 g
So, the maximum mass of carbon dioxide that could be produced by the chemical reaction is 17.6 g.