When 0.954 g of CaO is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 391C 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)
We will be happy to critique your thinking. We are not inclined to do it for you.
I wonder if you have omitted a decimal point in the T increase. An increase of 391 C from almost any starting point leaves us with superheated steam and the heat capacity of steam is not the same as that of water.
To determine the enthalpy change (ΔHrxn) for the reaction, we can use the formula:
ΔHrxn = q / n
Where:
ΔHrxn = Enthalpy change of the reaction
q = Amount of heat absorbed or released
n = Number of moles of the limiting reactant
First, we need to calculate the amount of heat absorbed or released (q) in the reaction.
q = mcΔT
Where:
q = Amount of heat absorbed or released
m = Mass of the solution (in grams)
c = Specific heat capacity of the solution
ΔT = Temperature change
Given:
Mass of the solution (m) = 200.0 mL x 1.00 g/mL (density) = 200.0 g
Specific heat capacity (c) = 4.184 J/g°C
Temperature change (ΔT) = 39.1 °C
Now, let's calculate q:
q = (200.0 g)(4.184 J/g°C)(39.1 °C)
q = 326,206.4 J
Next, we need to determine the number of moles (n) of the limiting reactant.
The balanced equation tells us that 1 mole of CaO reacts with 2 moles of H+.
Given:
Mass of CaO = 0.954 g
Molar mass of CaO = 56.08 g/mol
n = (0.954 g) / (56.08 g/mol)
n = 0.017 moles
Now we have all the necessary information to calculate ΔHrxn:
ΔHrxn = q / n
ΔHrxn = 326,206.4 J / 0.017 moles
Calculating the final answer gives the enthalpy change of the reaction.