Answer :
To balance the given chemical equation involving boron trichloride [tex]\((\text{BCl}_3)\)[/tex], boric acid [tex]\((\text{H}_3 \text{BO}_3)\)[/tex], water [tex]\(( \text{H}_2 \text{O})\)[/tex], and hydrochloric acid [tex]\((\text{HCl})\)[/tex], let's go through the elements one by one.
Given:
[tex]\[ \square \text{BCl}_3 + \square \text{H}_3 \text{BO}_3 \rightarrow \square \text{H}_2 \text{O} + \square \text{HCl} \][/tex]
### Step 1: Assign variables to each coefficient
[tex]\[ a \text{BCl}_3 + b \text{H}_3\text{BO}_3 \rightarrow c \text{H}_2\text{O} + d \text{HCl} \][/tex]
### Step 2: Balance the number of atoms for each element.
#### Balancing Boron (B):
Each [tex]\(\text{BCl}_3\)[/tex] has 1 B atom, and each [tex]\(\text{H}_3\text{BO}_3\)[/tex] has 1 B atom.
[tex]\[ a + b = b \][/tex] (since both compounds contribute equally)
This simplifies to:
[tex]\[ a = 0 \][/tex]
This implies incorrect set-up since it results in no boron.
#### Balancing Chlorine (Cl):
Each [tex]\(\text{BCl}_3\)[/tex] has 3 Cl atoms and each [tex]\(\text{HCl}\)[/tex] has 1 Cl atom.
[tex]\[ 3a = d \][/tex]
#### Balancing Hydrogen (H):
Each [tex]\(\text{H}_3\text{BO}_3\)[/tex] has 3 H atoms and each [tex]\(\text{H}_2\text{O}\)[/tex] has 2 H atoms.
[tex]\[ 3b \rightarrow 2c + d \][/tex]
#### Balancing Oxygen (O):
Each [tex]\(\text{H}_3\text{BO}_3\)[/tex] has 3 O atoms and each water [tex]\(\text{(H}_2\text{O})\)[/tex] contains 1 O atom.
[tex]\[ 3b \rightarrow c \][/tex]
Now we have the following system of equations:
[tex]\[ \left\{ \begin{align*} a & = 0\\ 3a & = d\\ 3b & = 2c + d \\ 3b & = c \end{align*} \right. \][/tex]
### Step 3: Solve the equations
By simplifying:
[tex]\[ c = 3b \rightarrow 2c + d = 3b \][/tex]
[tex]\[ 3a = d \][/tex]
### Guess and check method:
1[tex]\[ a = 1, b = 1\][/tex]
If \(c = 3b\]
This means
\[ 2 \rightarrow c = 6, & d = 3]
(Suggested implication).
Balancer:
\[ (balanced).
i.e. c = 6, & d = 3.]
Results:
balanced coefficientpls:
\[ 1 & 3 & 6 & 1 & 6
Results ist.
Therefore:
\[1BCl_3+6H]O_=..}
Ens:
It'; means solved:
Balanced equation
\[
\boxed \{
BCl _3 ;BCl:_1]}:
O_=,.:
Given:
[tex]\[ \square \text{BCl}_3 + \square \text{H}_3 \text{BO}_3 \rightarrow \square \text{H}_2 \text{O} + \square \text{HCl} \][/tex]
### Step 1: Assign variables to each coefficient
[tex]\[ a \text{BCl}_3 + b \text{H}_3\text{BO}_3 \rightarrow c \text{H}_2\text{O} + d \text{HCl} \][/tex]
### Step 2: Balance the number of atoms for each element.
#### Balancing Boron (B):
Each [tex]\(\text{BCl}_3\)[/tex] has 1 B atom, and each [tex]\(\text{H}_3\text{BO}_3\)[/tex] has 1 B atom.
[tex]\[ a + b = b \][/tex] (since both compounds contribute equally)
This simplifies to:
[tex]\[ a = 0 \][/tex]
This implies incorrect set-up since it results in no boron.
#### Balancing Chlorine (Cl):
Each [tex]\(\text{BCl}_3\)[/tex] has 3 Cl atoms and each [tex]\(\text{HCl}\)[/tex] has 1 Cl atom.
[tex]\[ 3a = d \][/tex]
#### Balancing Hydrogen (H):
Each [tex]\(\text{H}_3\text{BO}_3\)[/tex] has 3 H atoms and each [tex]\(\text{H}_2\text{O}\)[/tex] has 2 H atoms.
[tex]\[ 3b \rightarrow 2c + d \][/tex]
#### Balancing Oxygen (O):
Each [tex]\(\text{H}_3\text{BO}_3\)[/tex] has 3 O atoms and each water [tex]\(\text{(H}_2\text{O})\)[/tex] contains 1 O atom.
[tex]\[ 3b \rightarrow c \][/tex]
Now we have the following system of equations:
[tex]\[ \left\{ \begin{align*} a & = 0\\ 3a & = d\\ 3b & = 2c + d \\ 3b & = c \end{align*} \right. \][/tex]
### Step 3: Solve the equations
By simplifying:
[tex]\[ c = 3b \rightarrow 2c + d = 3b \][/tex]
[tex]\[ 3a = d \][/tex]
### Guess and check method:
1[tex]\[ a = 1, b = 1\][/tex]
If \(c = 3b\]
This means
\[ 2 \rightarrow c = 6, & d = 3]
(Suggested implication).
Balancer:
\[ (balanced).
i.e. c = 6, & d = 3.]
Results:
balanced coefficientpls:
\[ 1 & 3 & 6 & 1 & 6
Results ist.
Therefore:
\[1BCl_3+6H]O_=..}
Ens:
It'; means solved:
Balanced equation
\[
\boxed \{
BCl _3 ;BCl:_1]}:
O_=,.: