What are the products of the neutralization reaction of KOH with HCL

The products are KCl (potassium chloride) and H20 (water)

Explanation:

Hydrochloric acid (strong acid) reacts completely with potassium hydroxide (strong base) in a neutralization reaction (also called acid-base), giving as products: water and a salt (potassium chloride).

HCl + KOH---> KCl + H20

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

the answer is learned