Question

Balance the following equation using algebraic equilibrium.
Al2O3+HCL_ AL2CL3+H2O

Answer 1

To balance the equation \( \text{Al}_2\text{O}_3 + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2\text{O} \) using algebraic methods, we will assign variables to each chemical species.

Let:
- \( a \) = coefficient for \( \text{Al}_2\text{O}_3 \)
- \( b \) = coefficient for \( \text{HCl} \)
- \( c \) = coefficient for \( \text{AlCl}_3 \)
- \( d \) = coefficient for \( \text{H}_2\text{O} \)

The unbalanced equation can be expressed as:

\[
a \, \text{Al}_2\text{O}_3 + b \, \text{HCl} \rightarrow c \, \text{AlCl}_3 + d \, \text{H}_2\text{O}
\]

Now, let's set up the equations based on the conservation of each element:

1. **Aluminum (Al)**:
\[
2a = c \tag{1}
\]

2. **Chlorine (Cl)**:
\[
b = 3c \tag{2}
\]

3. **Oxygen (O)**:
\[
3a = d \tag{3}
\]

4. **Hydrogen (H)**:
\[
b = 2d \tag{4}
\]

Now we can solve these equations step by step.

From Equation (1):
\[
c = 2a
\]

Substituting \( c \) into Equation (2):
\[
b = 3(2a) = 6a
\]

From Equation (3):
\[
d = 3a
\]

Now substituting \( d \) into Equation (4):
\[
b = 2(3a) = 6a
\]

We see that our expressions hold consistently. Now we can simply choose a value for \( a \) to obtain the smallest whole number coefficients. Let’s assume \( a = 1 \):

Then:
- \( c = 2(1) = 2 \)
- \( b = 6(1) = 6 \)
- \( d = 3(1) = 3 \)

Thus, we have the coefficients:
- \( a = 1 \)
- \( b = 6 \)
- \( c = 2 \)
- \( d = 3 \)

Putting it back into the balanced equation:

\[
1 \, \text{Al}_2\text{O}_3 + 6 \, \text{HCl} \rightarrow 2 \, \text{AlCl}_3 + 3 \, \text{H}_2\text{O}
\]

So the balanced equation is:

\[
\text{Al}_2\text{O}_3 + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2\text{O}
\]