Balance the following equation using algebraic equilibrium.

Al2O3+HCL_ AL2CL3+H2O

1 answer

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:

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