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.
