Question

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

Explanation:

Step1: Balance Mg atoms

On the left, we have 1 Mg (from $\ce{MgCl_{2}}$), and on the right, we have 3 Mg (from $\ce{Mg_{3}(PO_{4})_{2}}$). So we put a coefficient of 3 in front of $\ce{MgCl_{2}}$. Now the equation is: $3\ce{MgCl_{2}} + \ce{K_{3}PO_{4}}
ightarrow \ce{Mg_{3}(PO_{4})_{2}} + \ce{KCl}$

Step2: Balance PO₄³⁻ ions

On the left, we have 1 $\ce{PO_{4}^{3-}}$ (from $\ce{K_{3}PO_{4}}$), and on the right, we have 2 $\ce{PO_{4}^{3-}}$ (from $\ce{Mg_{3}(PO_{4})_{2}}$). So we put a coefficient of 2 in front of $\ce{K_{3}PO_{4}}$. Now the equation is: $3\ce{MgCl_{2}} + 2\ce{K_{3}PO_{4}}
ightarrow \ce{Mg_{3}(PO_{4})_{2}} + \ce{KCl}$

Step3: Balance K atoms

On the left, we have 2×3 = 6 K (from $2\ce{K_{3}PO_{4}}$), and on the right, we have 1 K (from $\ce{KCl}$). So we put a coefficient of 6 in front of $\ce{KCl}$. Now the equation is: $3\ce{MgCl_{2}} + 2\ce{K_{3}PO_{4}}
ightarrow \ce{Mg_{3}(PO_{4})_{2}} + 6\ce{KCl}$

Step4: Balance Cl atoms (check)

On the left, we have 3×2 = 6 Cl (from $3\ce{MgCl_{2}}$), and on the right, we have 6 Cl (from $6\ce{KCl}$). All atoms are balanced now.

Answer:

$3\ce{MgCl_{2}} + 2\ce{K_{3}PO_{4}}
ightarrow \ce{Mg_{3}(PO_{4})_{2}} + 6\ce{KCl}$