
Answer:
There is a production of 11.6 moles of COâ‚‚
Explanation:
The reaction is this:
2C₂H₆(g)  +  7O₂(g)  ⟶  4CO₂(g)  +  6H₂O(g)
2 moles of ethane reacts with 7 moles of oxygen, to make 4 mol of dioxide and 6 moles of water vapor.
If the oxygen is in excess, we make the calculate with the ethane (limiting reactant)
2 moles of ethane produce 4 moles of dioxide
5.8 moles of ethane produce (5.8 Â .4)/2 = 11.6 moles