When 0.934 g of CaO is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 383C 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)
To determine the enthalpy change (ΔHrxn) for the given reaction, we can use the formula:
ΔHrxn = q / n
where q is the amount of heat absorbed or released during the reaction, and n is the number of moles of the limiting reactant.
First, let's find the number of moles of CaO added:
n(CaO) = mass / molar mass
n(CaO) = 0.934 g / 56.08 g/mol
n(CaO) = 0.0166 mol
Since the reaction equation shows that the ratio of CaO to H+ is 1:2, the number of moles of H+ is twice the number of moles of CaO.
n(H+) = 2 * n(CaO)
n(H+) = 2 * 0.0166 mol
n(H+) = 0.0332 mol
To calculate the heat absorbed or released during the reaction (q), we can use the formula:
q = m * C * ΔT
where m is the mass, C is the heat capacity, and ΔT is the temperature change.
First, let's find the mass of the solution:
mass = volume * density
mass = 200.0 mL * 1.00 g/mL
mass = 200.0 g
Using the given heat capacity (C = 4.184 J/g°C) and temperature change (ΔT = 38.3°C), we can now calculate the heat (q):
q = 200.0 g * 4.184 J/g°C * 38.3°C
q = 32040 J
Finally, we can calculate the enthalpy change (ΔHrxn):
ΔHrxn = q / n(H+)
ΔHrxn = 32040 J / 0.0332 mol
ΔHrxn = 965,060 J/mol
The enthalpy change (ΔHrxn) for the reaction is 965,060 J/mol.