Respuesta :
Answer:
All the ethane will be consumed by the chemical reaction, so no grams to react will be left over. This occurs because the ethanol is the limiting reagent.
Explanation:
Reactions that occur with a compound and oxygen are called combustion, and the products are always COâ‚‚ and water.
The equation for the combustion of ethane is:
2CH₃CH₃ (g) + 7O₂(g) →  4CO₂(g)  +  6H₂O(g)
We convert the mass of the reactants, in order to find out the moles of each.
2.1 g . 1 mol / 30 g = 0.07 moles
13.2 g / 32 g/mol = 0.412 moles
We now propose the following rule of three:
7 moles of oxgen need 2 moles of ethane to react
Then, 0.412 moles of Oâ‚‚ will react with (0.412 . 2) / 7 = 0.118 moles of ethane. We only have 0.07 moles, so we don't have enough ethane. That's why it is the limiting reactant.
All the ethane will be consumed by the chemical reaction, so no grams will be left over.
Answer:
Since ethane is the limiting reactant, it will completely be consumed. 0 grams will left over. O2 is in excess, there will remain 5.4 grams of oxygen
Explanation:
Step 1: Data given
Mass of ethane = 2.1 grams
Mass of oxygen = 13.2 grams
Molar mass of ethane = 30.07 g/mol
Molar mass of oxygen = 32.0 g/mol
Step 2: The balanced equation
2C2H6 + 7O2 → 4CO2 + 6H2O
Step 3: Calculate moles ethane
Moles ethane = 2.1 grams / 30.07 g/mol
Moles ethane = 0.0698 moles
Step 4: Calculate moles oxygen
Moles O2 = 13.2 grams / 32.0 g/mol
Moles O2 = 0.4125 moles
Step 5: Calculate the limiting reactant
For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
Ethane is the limiting reactant. It will completely be consumed (0.0698 moles). O2 is in excess. There will react 3.5 * 0.0698 = 0.2443 moles
There will remain 0.4125 - 0.2443 = 0.1682 moles
Step 6: calculate the mass oxygen remaining
Mass O2 = 0.1682 moles * 32.0 g/mol
Mass O2 = 5.4 grams
Since ethane is the limiting reactant, it will completely be consumed. 0 grams will left over. O2 is in excess, there will remain 5.4 grams of oxygen
