Questions: Solid potassium hydroxide (KOH) decomposes into gaseous water and solid potassium oxide. Write a balanced chemical equation for this reaction.

Solution Steps

The problem states that solid potassium hydroxide (KOH) decomposes into gaseous water (H₂O) and solid potassium oxide (K₂O).

First, write the unbalanced chemical equation using the chemical formulas of the reactants and products: \[ \mathrm{KOH} \rightarrow \mathrm{H_2O} + \mathrm{K_2O} \]

To balance the equation, ensure that the number of atoms of each element is the same on both sides of the equation.

1. Potassium (K): There are 2 K atoms on the product side (in K₂O), so we need 2 K atoms on the reactant side. Therefore, we place a coefficient of 2 in front of KOH: \[ 2 \mathrm{KOH} \rightarrow \mathrm{H_2O} + \mathrm{K_2O} \]

2. Oxygen (O): There are 2 O atoms on the reactant side (2 from 2 KOH) and 2 O atoms on the product side (1 from H₂O and 1 from K₂O). The oxygen atoms are balanced.

3. Hydrogen (H): There are 2 H atoms on the reactant side (2 from 2 KOH) and 2 H atoms on the product side (2 from H₂O). The hydrogen atoms are balanced.

The balanced chemical equation is: \[ 2 \mathrm{KOH} \rightarrow \mathrm{H_2O} + \mathrm{K_2O} \]

Final Answer