2. A 100 mL sample of 0.862 M HCI(aq) is mixed with 100 mL of 0.431 M Ba(OH)2(aq
ID: 591639 • Letter: 2
Question
2. A 100 mL sample of 0.862 M HCI(aq) is mixed with 100 mL of 0.431 M Ba(OH)2(aq) in a constant pressure calorimeter of negligible heat capacity. The initial temperature of HCl and Ba(OH)2 solutions is the same at 22.5 "C. The heat of neutralization for the process below is-56.2 kJ/mol. H+ (aq) + OH-(aq) H2O() What is the final temperature of the mixed solution? Assume the density of the solution is the same as that of water (1.00 g/mL), and the specific heat of the solution is the same as that for pure water, 4.184 J/(g.°C)Explanation / Answer
1 mol Ba(oh)2 = 2 mol HCl
no of mol of HCl reacted = 100*0.862/1000 = 0.0862 mol
no of mol of Ba(OH)2 reacted = 100*0.431/1000 = 0.0431 mol
heat liberated(q) = 56.2*0.0431 = 2.422 kj
q = m*s*DT
2.422*10^3 = 200*4.184*(x-22.5)
x = final temperature of solution = 25.4 c
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