Question

Attempt 3 During an experiment, a student adds 2.90 g CaO 2.90 g CaO to 400.0 mL 400.0 mL of 1.500 M HCl 1.500 M HCl . The student observes a temperature increase of 6.00 °C 6.00 °C . Assuming that the solution's final volume is 400.0 mL 400.0 mL , the density is 1.00 g / mL 1.00 g/mL , and the heat capacity is 4.184 J / g ⋅ °C 4.184 J/g⋅°C , calculate the heat of the reaction, Δ H rxn ΔHrxn . CaO ( s ) + 2 H + ( aq ) ⟶ Ca 2 + ( aq ) + H 2 O ( l )

Answer 1

Answer:

Explanation:

Equation of the reaction:

CaO(s) + 2H+(aq) -----> Ca2+(aq) + H2O(g)

The ∆Hrxn would be for one mole of CaO reacted or 2 moles of H+, whichever is the limiting reactant.

Number of moles = mass ÷ molar mass

Molar mass of CaO = 40 + 16

= 56 g/mol

moles of CaO = 2.90/56

= 0.0518 mol

Number of moles = concentration × volume

moles of HCl = 400 × 10^-3 × 1.500 = 0.6 moles

Moles of HCl = moles of H+

From the equation, 1 mole of CaO reacted with 2 moles of H+ to give 1 mole of water.

To find the limiting reagent,

0.6 mole of H+/2 moles of H+ × 1 mole of CaO

= 0.3 moles of CaO(> 0.0518 moles)

So, CaO is limiting reactant.

∆H = m × Cp × ∆T

m = density × volume

= 400 × 1

= 400 g

Cp = 4.184 J/g-ºC

∆T = +6 ºC

∆H = 400 × 4.184 × 6

= 10041.6 J

Since the reaction is exothermic,

∆Hrxn = -∆H/mol(CaO)

= -10041.6/0.0518

= -193853 J

= -193.9 kJ/mol.