2KOH + H2SO4 ==> 2H2O + K2SO4
mols KOH = M x L = 0.0217
mols H2SO4 = 0.01085
mols KOH needed = 2*0.01085 = 0.02170g so the solutions exactly neutralize each other.
mols H2O formed = 0.02170
volume = 21.70 + 21.70 = 43.40 mL
g H2O = 43.40 grams
qrxn = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = 43.40 g x 4.184 J/g*C x(30.17-23.50)
q = about 1210 J but you need a better number than this estimate.
About 1210 J for 0.0217 mol is
about 1210/0.0217 = about 55 kJ/mol
when 21.70 ml of 0.5000 m h2so4 is added to 21.70 mL of 1.000 M KOH in a coffee cup calorimeter at 23.50 degrees c, the temperature rises to 30.17 degree c. calculate delta H of this reaction (Assume that the total volume is the sum of the individual volumes and that density and specific heat capacity of the solution are the same as pure water d of water =1.00g/mL: c of water =4.184J/g*c) I have no idea where to even start
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