Consider the reaction: 2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = -118 kJ

Question

a) Calculate the heat (in kJ) released (make sure to use the correct sign in your answer) when 106.7 mL of 0.365 M HCl is mixed with excess Ba(OH)₂.

b) How many grams of barium hydroxide are required to produce enough energy to heat a 509.8 g of sample of water from 22.3°C to 96.6°C?

Answer 1

Ba(OH)₂ + 2HCl ==> BaCl₂ + 2H₂O ... balanced equation ... ∆Hrxn = -118 kJ

a) mol HCl = 0.1067 L x 0.365 mol/L = 0.03895 moles

∆H = 0.03895 mol HCl x -118 kJ/2 mol HCl = -2.30 kJ

b) First, we'll find the heat needed to heat 509.8 g of water from 22.3 to 96.6 degrees

q = mC∆T

q = (509.8 g)(4.184 J/gº)(74.3º) = 158,482 J = 158.5 kJ This is the heat needed

From the equation for the reaction of Ba(OH)₂ and HCl, we see that reaction of 1 mol Ba(OH)₂ will generate 118 kJ of heat. From this info, we can determine moles Ba(OH)₂ needed to generate 158.5 kJ.

158.5 kJ x 1 mol Ba(OH)₂ / 118 kJ = 1.35 moles Ba(OH)₂ needed

Converting this to grams, we have...

1.35 mol Ba(OH)₂ x 171.3 g/mol = 231 g Ba(OH)₂ needed