Question

directions: balance each chemical reaction below and then classify the chemical reactions as synthesis (s), decomposition (d), single-displacement (sd), double-displacement (dd), combustion (c), or acid-base (ab).

1. hbr + mg(oh)₂ → mgbr₂ + h₂o =
2. pbbr₂ + hcl → hbr + pbcl₂ =
3. cobr₃ + caso₄ → cabr₂ + co₂(so₄)₃ =
4. c₆h₁₂ + o₂ → co₂ + h₂o =
5. b₂o₃ + h₂o → h₃bo₃ =
6. h₂o → h₂ + o₂ =
7. c₂h₄ + o₂ → co₂ + h₂o =
8. mg + fe₂o₃ → mgo + fe =
9. h₃po₄ + ca(oh)₂ → ca₃(po₄)₂ + h₂o =
10. cl₂ + kbr → kcl + br₂ =
11. h₃aso₄ → as₂o₅ + h₂o =

12 s₃ + o₂ → so₃ =

1. bi(no₃)₃ + al₂(so₄)₃ → bi₂(so₄)₃ + al(no₃)₃ =
2. hgo → hg + o₂ =
3. p + o₂ → p₂o₃ =
4. hbr + ba(oh)₂ → babr₂ + h₂o =
5. fe + h₂o → fe₃o₄ + h₂ =

Explanation:

Step1: Balance the chemical equation

For \( \text{HBr}+\text{Mg(OH)}_2
ightarrow\text{MgBr}_2 + \text{H}_2\text{O} \), we use the method of balancing atoms.

• Br atoms: There is 1 Br on the left (in \( \text{HBr} \)) and 2 Br on the right (in \( \text{MgBr}_2 \)). So we put a coefficient of 2 in front of \( \text{HBr} \).
• H and O atoms: After putting 2 in front of \( \text{HBr} \), we have \( 2\text{HBr}+\text{Mg(OH)}_2

ightarrow\text{MgBr}_2 + \text{H}_2\text{O} \). Now, for H atoms: on the left, \( 2\text{HBr} \) gives 2 H and \( \text{Mg(OH)}_2 \) gives 2 H (from \( 2\text{OH} \)), total 4 H. On the right, \( \text{H}_2\text{O} \) has 2 H. So we put a coefficient of 2 in front of \( \text{H}_2\text{O} \).
The balanced equation is \( 2\text{HBr}+\text{Mg(OH)}_2=\text{MgBr}_2 + 2\text{H}_2\text{O} \).

Step2: Classify the reaction

This is an acid - base reaction (\( \text{AB} \)) because \( \text{HBr} \) (an acid) reacts with \( \text{Mg(OH)}_2 \) (a base) to form a salt (\( \text{MgBr}_2 \)) and water (\( \text{H}_2\text{O} \)).

Answer:

Balanced equation: \( 2\text{HBr}+\text{Mg(OH)}_2=\text{MgBr}_2 + 2\text{H}_2\text{O} \), Classification: Acid - Base (AB)