The chemical reaction for this is:
2 C2H6 + 7 O2 => 4 CO2 + 6 H2O
Solving for CO2 with each reactant will give:
21.0 g C2H6 x (1 mol C2H6/30.08 g C2H6) x (6 mol H2O/2
mol C2H6) x (18 g H2O/1 mol H2O) = 37.70 g H2O
110 g O2 x (1 mol O2/32.00 g O2) x (6 mol CO2/7 mol O2) x
(18 g H2O/1 mol H2O) = 53.04 g H2O
Since the amount of H2O in C2H6 is lower therefore C2H6
is the limiting reactant and the maximum amount of water is only 38 g H2O (2 significant digits)
ANswer:
38 g water
The reactions are a bit poorly written. While it's true that aqueous H₂CO₃ is produced in this neutralization reaction, the H₂CO₃ rapidly decomposes to yield CO₂(g) and H₂O(l). Writing the product as H₂CO₃(aq) in the net ionic equation is unnecessarily confusing since it portrays the substance as nonionizing yet water-soluble.
In any case, the Na⁺ and the Cl⁻ are the spectator ions here.
I believe it is b. I hope this helps you
Answer:
3966.82 J
Explanation:
q=sm∆T
q=73×13×4.18
the specific heat for water is 4.18
