Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 16.2 g of ethane is mixed with 19. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
This is a limiting reagent (LR) problem; we know that because amounts are given for BOTH reactants. I work these the long way because it easier to explain.
2C2H6 + 7O2 ==> 4CO2 + 6H2O
mols C2H6 = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols C2H6 to mols CO2.
Do the same for mols O2 to mols CO2.
It is likely these two values will not be the same which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that is the LR.
Using the smaller value, convert that to g CO2. g = mols CO2 x molar mass CO2.
To find the maximum mass of carbon dioxide produced, we need to determine the limiting reactant first. The reactant that is completely consumed will determine the amount of product formed.
1. Calculate the molar masses of ethane (C2H6) and oxygen (O2):
- Molar mass of ethane (C2H6):
C: 12.01 g/mol x 2 = 24.02 g/mol
H: 1.008 g/mol x 6 = 6.048 g/mol
Total: 24.02 g/mol + 6.048 g/mol = 30.068 g/mol
- Molar mass of oxygen (O2):
O: 16.00 g/mol x 2 = 32.00 g/mol
2. Convert the given masses of ethane and oxygen to moles:
- Moles of ethane:
Moles = Mass / Molar mass = 16.2 g / 30.068 g/mol ≈ 0.5392 mol
- Moles of oxygen:
Moles = Mass / Molar mass = 19.0 g / 32.00 g/mol ≈ 0.5938 mol
3. Determine the molar ratio between the reactants and product:
- Balanced chemical equation: 2C2H6 + 7O2 -> 4CO2 + 6H2O
From the balanced equation, we can see that 2 moles of ethane produce 4 moles of carbon dioxide.
4. Find the limiting reactant:
To find the limiting reactant, we compare the mole ratio of the reactants to the mole amounts given.
- Ethane: 0.5392 mol x (4 mol CO2 / 2 mol C2H6) = 1.0784 mol CO2
- Oxygen: 0.5938 mol x (4 mol CO2 / 7 mol O2) = 0.3392 mol CO2
The limiting reactant is oxygen since it produces fewer moles of carbon dioxide.
5. Calculate the maximum mass of carbon dioxide produced:
- Moles of carbon dioxide produced will be equal to the moles of oxygen consumed: 0.3392 mol CO2
- Mass of carbon dioxide produced:
Mass = Moles x Molar mass = 0.3392 mol x 44.01 g/mol ≈ 14.94 g
Rounded to the correct number of significant digits, the maximum mass of carbon dioxide produced is approximately 14.9 g.
To calculate the maximum mass of carbon dioxide produced, we need to determine which reactant limits the reaction and calculate the amount of carbon dioxide produced from that reactant.
First, let's calculate the molar masses of the compounds involved:
- Molar mass of ethane (C2H6): 2 * atomic mass of carbon + 6 * atomic mass of hydrogen
- Molar mass of oxygen (O2): 2 * atomic mass of oxygen
- Molar mass of carbon dioxide (CO2): atomic mass of carbon + 2 * atomic mass of oxygen
Using the atomic masses from the periodic table:
- Molar mass of ethane (C2H6): 2 * 12.01 g/mol + 6 * 1.01 g/mol = 30.07 g/mol
- Molar mass of oxygen (O2): 2 * 16.00 g/mol = 32.00 g/mol
- Molar mass of carbon dioxide (CO2): 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol
Next, we need to determine the limiting reactant by comparing the moles of ethane and oxygen.
- Moles of ethane = mass of ethane / molar mass of ethane
- Moles of oxygen = mass of oxygen / molar mass of oxygen
- Moles of ethane = 16.2 g / 30.07 g/mol = 0.539 mol
- Moles of oxygen = 19. g / 32.00 g/mol = 0.594 mol
To determine the limiting reactant, we compare the mole ratios between ethane and oxygen in the balanced chemical equation. The balanced equation for the reaction is:
C2H6 + 3.5 O2 → 2 CO2 + 3 H2O
From the balanced equation, the mole ratio between ethane and oxygen is 1:3.5.
Since the moles of oxygen (0.594 mol) are greater than the ratio of ethane (0.539 mol), oxygen is in excess and ethane is the limiting reactant.
Next, we need to calculate the moles of carbon dioxide produced from the limiting reactant (ethane) using the mole ratio from the balanced equation:
- Moles of carbon dioxide = Moles of ethane * (2 moles of CO2 / 1 mole of C2H6)
- Moles of carbon dioxide = 0.539 mol * (2 mol CO2 / 1 mol C2H6) = 1.078 mol CO2
Finally, we can calculate the maximum mass of carbon dioxide produced using the mole ratio and the molar mass of carbon dioxide:
- Mass of carbon dioxide = Moles of carbon dioxide * Molar mass of CO2
- Mass of carbon dioxide = 1.078 mol * 44.01 g/mol = 47.361 g
Rounding to the correct number of significant digits (based on the given data), the maximum mass of carbon dioxide produced is 47.4 g (3 significant digits).