6. balance the following neutralization reactions:
a) hbr + koh → kbr + h₂o
b) hf + fe(oh)₃ → fef₃ + h₂o
c) hcl + ca(oh)₂ → cacl₂ + h₂o

Question

Accepted Answer

# Explanation:
## Step1: Balance equation a
- **Balance $H$ and $O$**:
  - For $HBr + KOH
ightarrow KBr + H_{2}O$, count the number of atoms. On the left - hand side, $H$ atoms: $1$ (from $HBr$) $+ 1$ (from $KOH$) $=2$, $Br$: $1$, $K$: $1$, $O$: $1$. On the right - hand side, $H$: $2$ (from $H_{2}O$), $Br$: $1$, $K$: $1$, $O$: $1$. So the balanced equation is $HBr+KOH = KBr + H_{2}O$, and the coefficients are $1$, $1$, $1$, $1$.
## Step2: Balance equation b
- **Balance $F$ and $H$**:
  - For $HF+Fe(OH)_{3}
ightarrow FeF_{3}+H_{2}O$, to balance $F$ atoms, since there are $3$ $F$ atoms in $FeF_{3}$, we need $3$ $HF$. Then for $H$ and $O$ atoms:
    - The left - hand side has $3$ $H$ (from $HF$) $+3$ $H$ (from $Fe(OH)_{3}$) $=6$ $H$ and $3$ $O$. The right - hand side, if we have $3$ $H_{2}O$, there are $6$ $H$ and $3$ $O$. So the balanced equation is $3HF + Fe(OH)_{3}=FeF_{3}+3H_{2}O$, and the coefficients are $3$, $1$, $1$, $3$.
## Step3: Balance equation c
- **Balance $Cl$ and $H$**:
  - For $HCl+Ca(OH)_{2}
ightarrow CaCl_{2}+H_{2}O$, to balance $Cl$ atoms (there are $2$ $Cl$ in $CaCl_{2}$), we need $2$ $HCl$. Then for $H$ and $O$ atoms:
    - The left - hand side has $2$ $H$ (from $HCl$) $+2$ $H$ (from $Ca(OH)_{2}$) $=4$ $H$ and $2$ $O$. The right - hand side, if we have $2$ $H_{2}O$, there are $4$ $H$ and $2$ $O$. So the balanced equation is $2HCl+Ca(OH)_{2}=CaCl_{2}+2H_{2}O$, and the coefficients are $2$, $1$, $1$, $2$.

# Answer:
a) $1$ $HBr+1$ $KOH
ightarrow1$ $KBr + 1$ $H_{2}O$
b) $3$ $HF+1$ $Fe(OH)_{3}
ightarrow1$ $FeF_{3}+3$ $H_{2}O$
c) $2$ $HCl + 1$ $Ca(OH)_{2}
ightarrow1$ $CaCl_{2}+2$ $H_{2}O$

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QUESTION IMAGE

Question

Explanation:

Step1: Balance equation a

Step2: Balance equation b

Step3: Balance equation c

Answer: