q = mass x specific heat x (Tfinal-Tinitial) = approx 1.1 kJ but you should get a better answer than this estimate. That is for approx 0.02 mols H2O produced. You should confirm all of those estimates.
Then kJ/mol = 1.1/0.02 = ? kJ/mol.
Note: Where did the 0.02 mols H2O produced come fromj?
You had mols H2SO4 = M x L = about 56.
You had M x L NaOH = about 20 so NaOH is the limiting reagent. Go through the equation to confirm that.
In a constant-pressure calorimeter, 60.0 mL of 0.930 M H2SO4 was added to 60.0 mL of 0.320 M NaOH. The reaction caused the temperature of the solution to rise from 22.68 °C to 24.86 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes. answer in KJ/mol of H2O
1 answer