Respuesta :
Answer:
Ethane is the limiting reactant, so no mass of ethane will be left over by the chemical reaction. All the mass will react, the 2.1 grams of ethane.
Explanation:
First of all, we need to determine the reaction and the limiting reactant to work with the stoichiometry.
The equation is: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
We define the moles of the reactants:
2.10 g / 30 g/mol = 0.07 moles of ethane
12 g / 32 g/mol = 0.375 moles of oxygen
To determine the limiting reactant, we start with oxygen:
7 moles of O₂ can react with 2 moles of ethane
Then, 0.375 moles of O₂ will react with (0.375 . 2) / 7= 1.31 moles of ethane.
We do not have enough ethane, just only 0.07 moles to react.
Ethane is the limiting reactant, so no mass of ethane will be left over by the chemical reaction. All the mass will react, the 2.1 grams of ethane.
Answer:
Since ethane is the limting reactant, there will remain nothing . The mass ethane remaingin is 0 grams
There remains 4.18 grams of oxygen
Explanation:
Step 1: data given
Ethane = C2H6
Mass of ethane = 2.10 grams
Molar mass C2H6 = 30.07 g/mol
oxygen = O2
Mass of oxygen = 12.0 grams
Molar mass O2 = 32.0 g/mol
Step 2: The balanced equation
2C2H6 + 7O2 → 4CO2 + 6H2O
Step 3: Calculate moles C2H6
Moles C2H6 = mass C2H6 / molar mass C2H6
Moles C2H6 = 2.10 grams / 30.07 g/mol
Moles C2H6 = 0.0698 moles
Step 4: Calculate moles O2
Moles O2 = 12.0 grams / 32.0 g/mol
Moles O2 = 0.375 moles
Step 5: Calculate limiting reactant
For 2 moles ethane we have 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
Ethane is the limiting reactant. It wil completely be consumed (0.0698 moles). O2 is in excess. There will react 3.5 * 0.0698 = 0.2443 moles
There will remain 0.375 - 0.2443 = 0.1307 moles
Since ethane is the limting reactant, there will remain nothing . The mass ethane remaingin is 0 grams
Step 6: Calculate mass oxygen remaining
Mass oxygen = 0.1307 moles * 32.0 g/mol
Mass oxygen remaining = 4.18 grams
