50 ml 1.0 M HCl mixed with 50.0 ml 1.0 M NaOH please shoe solution full equation

Question

50 ml 1.0 M HCl mixed with 50.0 ml 1.0 M NaOH

please shoe solution

full equation HCl_(aq) + NaOH_(aq) rightarrow NaCl_(aq) + H_2O net ionic equation H^+ + OH^- rightarrow H_2O volume of acid solution _____ mL temperature _____ degree C volume of base solution _____ mL temperature _____ degree C total volume _____ mL average temp. _____ degree C high temperature after reaction _____ degree C change in temperature _____ degree C Heat Gained by Water _____ J Gained by Calorimeter _____ J Heat Released by Reaction _____ J Moles of reactant _____ moles Heat of Reaction _____ J/mole

Explanation / Answer

Heat gained by water = 100 x 4.18 x (29.7 - 23.1) = 2758.8 J

Heat gained by calorimeter = 0 (since heat capacity of calorimeter is not provided)

Heat release by reaction = Heat gained by water = 2758.8 J

Moles of reactant = 1 x 0.05 = 0.05 moles of HCl and  x 0.05 = 0.05 moles of NaOH

2758.8 J is for 0.05 moles of reactant

so for 1 mole of reactant => 2758.8/.05 = 55176 J

so Heat of Reaction = 55176 J/mole

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