Question

Solid potassium hydroxide koh decomposes into gaseous water and solid potassium oxide . write a balanced chemical equation for this reaction.

Answer 1

The balanced chemical equation for the decomposition of solid potassium hydroxide (KOH) into solid potassium oxide (K2O) and gaseous water (H2O) is: 4 KOH(s) → 2 K2O(s) + 2 H2O(g).

Step-by-step explanation:

The balanced chemical equation for the decomposition of solid potassium hydroxide (KOH) into gaseous water (H2O) and solid potassium oxide (K2O) is:

4 KOH(s) → 2 K2O(s) + 2 H2O(g)

This reaction showcases the breakdown of potassium hydroxide into simpler substances when it decomposes. Notice that the total number of atoms for each element is conserved on both sides of the equation, fulfilling the Law of Conservation of Mass.

Answer 2

2KOH(s) ---> K2O(s) + H2O(g)

Step-by-step explanation:

potassium hydroxide = KOH

water = H2O

potassium oxide = K2O

since KOH decomposes into H2O and K2O there is an arrow after KOH then after that you balance it. Without doing anything, KOH ----> H2O + K2O on the left side of the arrow there is only one K, one O, and one H atoms while on the other side there are 2 H, 2O, and 2K which means we have to put the two in front of the KOH. so then you will have 2K, 2O, 2H on the left-handed side which will equal the number of atoms there are on the right-handed side.

Also dont forget to put whether it is g, s, or aq.

In the instructions it tells you which one is s , g, or aq.