Question

Задача.К раствору ,содержащему 6,3 г. гидрооксида калия прилили р-р 6,3 г. азотной кислоты.Какое вещество останется не прореагировавшем полностью. Задание 2. Написать 2 реакции нейтрализации.

Answer 1

NKOH = 6,3/56 = 0,1125mol
nHNO3 = 6,3/63 = 0,1mol

Answer 2

Task 1: Excess Reactant in the Solution

When 6.3 g of potassium hydroxide is added to a solution containing 6.3 g of nitric acid, we can determine the excess reactant by comparing the stoichiometry of the reaction. The balanced chemical equation for the reaction between potassium hydroxide (KOH) and nitric acid (HNO3) is:

2KOH + HNO3 → KNO3 + H2O

From the balanced equation, we can see that the molar ratio of KOH to HNO3 is 2:1. To find the limiting reactant and the excess reactant, we need to compare the moles of each reactant.

First, let's calculate the moles of each substance: - Moles of KOH = Mass of KOH / Molar mass of KOH - Moles of HNO3 = Mass of HNO3 / Molar mass of HNO3

The substance that is not fully consumed in the reaction will be the excess reactant.

Task 2: Neutralization Reactions

Two examples of neutralization reactions are: 1. Hydrochloric Acid and Sodium Hydroxide: - The balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is: HCl + NaOH → NaCl + H2O

2. Sulfuric Acid and Potassium Hydroxide: - The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) is: H2SO4 + 2KOH → K2SO4 + 2H2O

These reactions are examples of acid-base neutralization reactions, where an acid reacts with a base to form a salt and water.

I hope this helps! If you have any further questions, feel free to ask.