Question

mg(oh)₂ + hcl → mgcl₂ + h₂o

Explanation:

Step1: Balance Mg atoms

On left: 1 Mg (in $\ce{Mg(OH)2}$), on right: 1 Mg (in $\ce{MgCl2}$). Mg is balanced.

Step2: Balance O and H from $\ce{OH^-}$

$\ce{Mg(OH)2}$ has 2 $\ce{OH^-}$ groups, so we need 2 $\ce{H2O}$ to balance O and H from $\ce{OH^-}$. Now, right side has 2 $\ce{H2O}$, so H on right: 4 (from 2 $\ce{H2O}$), O on right: 2 (from 2 $\ce{H2O}$).

Step3: Balance H from HCl

Left side H from HCl: each HCl gives 1 H. We need 4 H, but wait, no—wait, $\ce{Mg(OH)2 + HCl -> MgCl2 + H2O}$. Wait, $\ce{OH^-}$ (2 of them) react with $\ce{H^+}$ to form $\ce{H2O}$. So 2 $\ce{OH^-}$ need 2 $\ce{H^+}$? No, wait, $\ce{Mg(OH)2}$ is a base, HCl is acid. The reaction is neutralization. Let's do it properly.

$\ce{Mg(OH)2 + HCl -> MgCl2 + H2O}$

• Mg: 1 on left, 1 on right. Good.
• O: 2 on left (from $\ce{Mg(OH)2}$), 1 on right (from $\ce{H2O}$). So need 2 $\ce{H2O}$ to balance O: O becomes 2 on right.
• H: Left: 2 (from $\ce{Mg(OH)2}$) + x (from HCl). Right: 22 = 4 (from 2 $\ce{H2O}$). So 2 + x = 4 → x=2? No, wait, HCl has H and Cl. Wait, Cl: Left: x (from HCl), Right: 2 (from $\ce{MgCl2}$). So x=2. Wait, but then H: Left: 2 (from $\ce{Mg(OH)2}$) + 2 (from 2 HCl) = 4. Right: 22 = 4 (from 2 $\ce{H2O}$). O: Left: 2, Right: 2 (from 2 $\ce{H2O}$). Mg: 1,1. Cl: 2,2. So the balanced equation is $\ce{Mg(OH)2 + 2 HCl -> MgCl2 + 2 H2O}$. Wait, let's check again.

Wait, original equation: $\ce{Mg(OH)2 + HCl -> MgCl2 + H2O}$

Let's list atoms:

• Mg: 1 (left) vs 1 (right) → balanced.
• O: 2 (left, $\ce{Mg(OH)2}$) vs 1 (right, $\ce{H2O}$) → need 2 $\ce{H2O}$ → O: 2 (right).
• H: Left: 2 (from $\ce{Mg(OH)2}$) + n (from HCl) → Right: 2*2 = 4 (from 2 $\ce{H2O}$). So 2 + n = 4 → n=2. So HCl: 2 moles.
• Cl: Left: n (from 2 HCl) = 2, Right: 2 (from $\ce{MgCl2}$) → balanced.

So the balanced equation is $\ce{Mg(OH)2 + 2 HCl = MgCl2 + 2 H2O}$.

Wait, let's do step by step:

1. Start with $\ce{Mg(OH)2 + HCl -> MgCl2 + H2O}$

1. Balance Mg: already 1 on each side.

1. Balance O: $\ce{Mg(OH)2}$ has 2 O, so put 2 in front of $\ce{H2O}$: $\ce{Mg(OH)2 + HCl -> MgCl2 + 2 H2O}$

1. Balance H: $\ce{2 H2O}$ has 4 H. $\ce{Mg(OH)2}$ has 2 H, so HCl needs to provide 2 H. But HCl is $\ce{HCl}$, so each HCl provides 1 H. So need 2 HCl: $\ce{Mg(OH)2 + 2 HCl -> MgCl2 + 2 H2O}$

1. Check Cl: 2 HCl gives 2 Cl, $\ce{MgCl2}$ has 2 Cl. Balanced.

1. Check all atoms:

• Mg: 1,1 ✔️

• O: 2,2 ✔️

• H: 2 + 2 = 4; 2*2 = 4 ✔️

• Cl: 2,2 ✔️

So the balanced equation is $\ce{Mg(OH)2 + 2 HCl = MgCl2 + 2 H2O}$.

Answer:

The balanced chemical equation is $\boldsymbol{\ce{Mg(OH)2 + 2 HCl -> MgCl2 + 2 H2O}}$