2KOH + H2SO4 ==> H2SO4 + 2H2O
q = -56 kJ/mol KOH or 2-56 = -112 kJ/rxn.
50.0 mL x 0.500M KOH = 25 mmols.
25.0 mL x 0.500M H2SO4 = 12.5 mmols.
Total volume = 75 mL and I assume density is considered to be 1.00 g/mL.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
112 = 75g x 4.184 J/g*K x (Tfinal-22.8)
Solve for Tf
I get an estimate of 23+ C
please help solve this problem now. I really need an answer. thank you
When one mole of KOH is neutralized by sulfuric acid, q=-56kJ. At 22.8 C , 25.0 mL of 0.500 M H2SO4 is neutralized by 50.0 mL of 0.500 M KOH in a coffee-cup calorimeter. What is the final temperature of the solution?
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