Question

balance the equation
mgcl₂ + k₃po₄ → mg₃(po₄)₂ + kcl

Explanation:

Step1: Balance Mg atoms

On the left, we have 1 Mg (from \( \text{MgCl}_2 \)), and on the right, we have 3 Mg (from \( \text{Mg}_3(\text{PO}_4)_2 \)). So we put a coefficient of 3 in front of \( \text{MgCl}_2 \):
\( 3\text{MgCl}_2 + \text{K}_3\text{PO}_4
ightarrow \text{Mg}_3(\text{PO}_4)_2 + \text{KCl} \)

Step2: Balance PO₄³⁻ ions

On the left, we have 1 \( \text{PO}_4^{3-} \) (from \( \text{K}_3\text{PO}_4 \)), and on the right, we have 2 \( \text{PO}_4^{3-} \) (from \( \text{Mg}_3(\text{PO}_4)_2 \)). So we put a coefficient of 2 in front of \( \text{K}_3\text{PO}_4 \):
\( 3\text{MgCl}_2 + 2\text{K}_3\text{PO}_4
ightarrow \text{Mg}_3(\text{PO}_4)_2 + \text{KCl} \)

Step3: Balance K atoms

On the left, we have \( 2 \times 3 = 6 \) K (from \( 2\text{K}_3\text{PO}_4 \)), and on the right, we have 1 K (from \( \text{KCl} \)). So we put a coefficient of 6 in front of \( \text{KCl} \):
\( 3\text{MgCl}_2 + 2\text{K}_3\text{PO}_4
ightarrow \text{Mg}_3(\text{PO}_4)_2 + 6\text{KCl} \)

Step4: Balance Cl atoms (verify)

On the left, we have \( 3 \times 2 = 6 \) Cl (from \( 3\text{MgCl}_2 \)), and on the right, we have 6 Cl (from \( 6\text{KCl} \)). All atoms are now balanced.

Answer:

\( 3\text{MgCl}_2 + 2\text{K}_3\text{PO}_4
ightarrow \text{Mg}_3(\text{PO}_4)_2 + 6\text{KCl} \)