Write and balance a chemical equation for the reaction between hydrobromic acid and beryllium hydroxide.

Reactants: Hydrobromic acid, Beryllium hydroxide
Products: Beryllium bromide, Water

Word equation:

$$\text{Hydrobromic acid} + \text{Beryllium hydroxide} \rightarrow \text{Beryllium bromide} + \text{Water}$$

Hydrobromic acid: HBr
Beryllium hydroxide: Be(OH)$_2$
Beryllium bromide: BeBr$_2$
Water: H$_2$O

Skeleton equation:

$$\text{HBr} + \text{Be(OH)}_2 \rightarrow \text{BeBr}_2 + \text{H}_2\text{O}$$

Reactants:
H: 1 (from HBr) + 2 (from Be(OH)$_2$) = 3
Br: 1
Be: 1
O: 2

Products:
H: 2
Br: 2
Be: 1
O: 1

There are 2 Br on the product side, so place a coefficient of 2 in front of HBr.

$$2\text{HBr} + \text{Be(OH)}_2 \rightarrow \text{BeBr}_2 + \text{H}_2\text{O}$$

Reactants:
H: 2 (from 2HBr) + 2 (from Be(OH)$_2$) = 4
Br: 2
Be: 1
O: 2

Products:
H: 2
Br: 2
Be: 1
O: 1

There are 4 H and 2 O on the reactant side. To balance, place a coefficient of 2 in front of H$_2$O.

$$2\text{HBr} + \text{Be(OH)}_2 \rightarrow \text{BeBr}_2 + 2\text{H}_2\text{O}$$

Reactants:
H: 2 + 2 = 4
Br: 2
Be: 1
O: 2

Products:
H: 2 $\times$ 2 = 4
Br: 2
Be: 1
O: 2

All atoms are balanced.

Balanced equation:

$$\boxed{2\text{HBr} + \text{Be(OH)}_2 \rightarrow \text{BeBr}_2 + 2\text{H}_2\text{O}}$$