The answer is actually D. The pH of the products can be different from the pH of the reactants. Hope this helps :)
Answer:
10.6 moles of CO₂ are produced in this combustion
Explanation:
The combustion reaction is:
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (g)
We assume the ethane as the limiting reactant because the excersise states that the O₂ is in excess.
We make a rule of three:
2 moles of ethane can produce 4 moles of CO₂
Therefore 5.30 moles of ethane will produce (5.3 . 4) /2 = 10.6 moles
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