Questions: KOH + H3PO4 → K3PO4 + H2O

Transcript text: (3) $\mathrm{KOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow \mathrm{~K}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}$

Solution

Solution Steps

Step 1: Identify the Reactants and Products

The given chemical equation is: \[ \mathrm{KOH} + \mathrm{H}_{3}\mathrm{PO}_{4} \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4} + \mathrm{H}_{2}\mathrm{O} \] Here, the reactants are potassium hydroxide (KOH) and phosphoric acid ($\mathrm{H}_{3}\mathrm{PO}_{4}$). The products are potassium phosphate ($\mathrm{K}_{3}\mathrm{PO}_{4}$) and water ($\mathrm{H}_{2}\mathrm{O}$).

Step 2: Balance the Chemical Equation

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

Potassium (K): There are 3 K atoms in $\mathrm{K}_{3}\mathrm{PO}_{4}$, so we need 3 KOH molecules.
Phosphorus (P): There is 1 P atom in $\mathrm{H}_{3}\mathrm{PO}_{4}$ and 1 P atom in $\mathrm{K}_{3}\mathrm{PO}_{4}$, so the P atoms are already balanced.
Oxygen (O): There are 4 O atoms in $\mathrm{H}_{3}\mathrm{PO}_{4}$ and 4 O atoms in $\mathrm{K}_{3}\mathrm{PO}_{4}$. Additionally, there are 3 O atoms in 3 KOH and 1 O atom in $\mathrm{H}_{2}\mathrm{O}$. This gives a total of 7 O atoms on both sides.
Hydrogen (H): There are 3 H atoms in $\mathrm{H}_{3}\mathrm{PO}_{4}$ and 3 H atoms in 3 KOH, giving a total of 6 H atoms. On the product side, there are 2 H atoms in $\mathrm{H}_{2}\mathrm{O}$, so we need 3 $\mathrm{H}_{2}\mathrm{O}$ molecules to balance the H atoms.

The balanced equation is: \[ 3 \mathrm{KOH} + \mathrm{H}_{3}\mathrm{PO}_{4} \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4} + 3 \mathrm{H}_{2}\mathrm{O} \]