Question

In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temperature of the solution to rise from 23.31 ∘C to 28.08 ∘C.
If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g⋅°C,) respectively), what is Δ𝐻 for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
millimoles Ba(OH)2 = mL x M = 70.00 x 0.350 = 24.5
millimoles HCl ==> 70 x 0.700 = 49.00
You can see that there is exactly enough of each reagent to complete the reaction; i.e., there is no limiting reagent.Therefore, 49.00 mmols H2O will form.
delta T = 28.08 - 23.31 = 4.77 degrees C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = 140 g x 4.184 J/g*C x (4.77) = about 2794 J for the formation of
49 mmols or 0.049 mols H2O
dH = 2.794 kJ x 1 mol/0.049 = about 57 kJ/mol H2O
Since this is an exothermic reaction, that would be reported as dH = -57 kJ/mol.
You should go through each step, confirm the numbers because I've estimated and rounded here and there.