sulphuric acid is a diprotic acid so has two H+ and needs two moles of sulphuric acid to neutralise it
Yes, the reaction involving the solid is actually an individual step in the equation of the reaction between the solutions. If you were to add the change in enthalpy of the reaction with the solid NaOh to the change in enthalpy of the other step in the reaction (that's adding water and the NaOh pellets) you would find the sum equivalent to the change in enthalpy of the reaction involving the two solutions (this is supported by Hess's law). I suggest that you consider Hess's law for more information.
formic acid and hydrochloric aciod
Run
The neutralization reaction of sodium hydroxide and hydrochloric acid will form sodium chloride (common salt) and water. NaOH +HCl --> NaCl+ H2O
Neutralization
Yes, the reaction involving the solid is actually an individual step in the equation of the reaction between the solutions. If you were to add the change in enthalpy of the reaction with the solid NaOh to the change in enthalpy of the other step in the reaction (that's adding water and the NaOh pellets) you would find the sum equivalent to the change in enthalpy of the reaction involving the two solutions (this is supported by Hess's law). I suggest that you consider Hess's law for more information.
alkaline
formic acid and hydrochloric aciod
Run
The reaction is:HCl + NaOH = NaCl + H2O
The neutralization reaction of sodium hydroxide and hydrochloric acid will form sodium chloride (common salt) and water. NaOH +HCl --> NaCl+ H2O
Neutralization
Neutralization
Aluminum hydroxide plus hydrochloric acid equals( produces ) aluminum chloride plus water.
it is always water.
Any reaction between sodium chloride and hydrochloric acid.
Neutralization reaction occurs.