When a solution containing 8.00 g NaOH in 50.0 g of water at 25 °C is added to a solution of 8.00 g of HCl in 250.0 g of water at 25 °C in a calorimeter, the temperature of the solution increases to 33.5 °C. Assuming that the specific heat heat of the solution is 4.184 J/(g۰°C) and the calorimeter absorbs a negligible amount of heat, calculate ∆H in kJ for the reaction

Question

When a solution containing 8.00 g NaOH in 50.0 g of water at 25 °C is added to a solution of 8.00 g of HCl in 250.0 g of water at 25 °C in a calorimeter, the temperature of the solution increases to 33.5 °C.  Assuming that the specific heat heat of the solution is 4.184 J/(g۰°C) and the calorimeter absorbs a negligible amount of heat, calculate ∆H in kJ for the reaction
NaOH(aq) + HCl(aq) →NaCl(aq) + H2O(l)

DrBob222 · Answer

8g NaOH = 8/40 = 0.2 mols
8g HCl = 8/36.5 = 0.22 mol

So NaOH is the limiting reagent.
q = mass x secific heat H2O x (Tfinal-Tinitial)
q = (50+250) x 4.184 x (33.5-25) = ?
q is delta H in J (convert to kJ) and that is all the post asks for. Usually these problems want kJ/mol

Joe · Answer

Should be (8+8+50+250) x 4.184 x (33.5-25)=q

Anonymous · Answer

When a solution containing 8.00 g of NaOH in 50.0 g of water at 25.0 °C is added to asolution of 8.00 g of HCl in 250.0 g of water at 25.0 °C in a calorimeter, the temperature ofthe solution increases to 33.5 °C. Assuming that the specific heat of the solution is 4.184Jg ∙ °C and that of the calorimeter absorbs a negligible amount of heat, calculate ∆ H in kJfor the reactionNaOH(aq)+ HCl(aq) → NaCl(aq)+ H2 O(