When 0.911 g of CaO is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 374C is observed.
Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
Hrxn, for the reaction of
CaO(s) + 2H+(aq)--> Ca2+(aq) + H2O(l)
Variations of this problem have shown up at least a dozen times in the last three or four days. I don't know how the temperature of a water solution can increase by 374 degrees C. I suppose it might be technically feasible if we started at absolute zero (-273 degrees C) and moved to 101 degrees C but I think it's highly unlikely to happen. Even then we have steam at 101 and that's still a problem, at least I think it is. Perhaps I'm on the wrong track here. I'll be glad to listen.
My theory is that someone looked up delta H for the reaction, stuck in a number for the mass of CaO, (or Ca for some of the questions I've seen) and mass of water, and worked backwards to get a delta T value. Then that someone didn't think about the unrealistic values for delta T. But that's just my theory.
the temp sould be 37.4 C ...sorry bout that
Well, that shoots my theory all to H-ll. It's as simple as a typo. Thanks for letting me know.
To calculate the enthalpy change (ΔH) for the given reaction, we need to use the equation:
ΔHrxn = q / n
where ΔHrxn is the enthalpy change for the reaction, q is the heat energy absorbed or released by the reaction, and n is the number of moles of the limiting reactant.
First, let's calculate the number of moles of CaO (calcium oxide) and HCl (hydrochloric acid) used in the reaction.
Given:
Mass of CaO = 0.911 g
Molar mass of CaO = 56.08 g/mol
Using the equation:
moles of CaO = mass of CaO / molar mass of CaO
moles of CaO = 0.911 g / 56.08 g/mol
moles of CaO ≈ 0.01624 mol (rounded to five significant figures)
Next, let's calculate the moles of HCl present in the solution.
Given:
Volume of HCl = 200.0 mL
Molarity of HCl = 0.500 M
Using the equation:
moles of HCl = volume of HCl (in liters) * molarity of HCl
moles of HCl = 200.0 mL * (1 L / 1000 mL) * 0.500 M
moles of HCl = 0.1000 mol (rounded to four significant figures)
The balanced chemical equation shows that the stoichiometric ratio between CaO and HCl is 1:2. So, the moles of HCl needed to react with all of the CaO will be twice the moles of CaO.
moles of HCl needed = 2 * moles of CaO
moles of HCl needed = 2 * 0.01624 mol
moles of HCl needed ≈ 0.03248 mol (rounded to five significant figures)
Now, we have the number of moles of the limiting reactant (HCl). Let's calculate the heat energy (q) absorbed or released by the reaction.
Given:
Temperature increase (ΔT) = 37.4 °C
Heat capacity = 4.184 J/g°C
Mass of the solution = volume of the solution * density of the solution
Mass of the solution = 200.0 mL * 1.00 g/mL
Mass of the solution = 200.0 g
Using the equation:
q = heat capacity * mass of the solution * ΔT
q = 4.184 J/g°C * 200.0 g * 37.4 °C
q ≈ 313,753.6 J (rounded to four significant figures)
Finally, let's calculate the enthalpy change (ΔH) for the reaction.
ΔHrxn = q / n
ΔHrxn = 313,753.6 J / 0.03248 mol
ΔHrxn ≈ 9,652,297 J/mol (rounded to four significant figures)
Therefore, the enthalpy change (ΔH) for the reaction of CaO(s) + 2H+(aq) → Ca2+(aq) + H2O(l) is approximately 9,652,297 J/mol.