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:
-
Aluminum (Al): \[ 2a = c \tag{1} \]
-
Chlorine (Cl): \[ b = 3c \tag{2} \]
-
Oxygen (O): \[ 3a = d \tag{3} \]
-
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:
\[ \text{Al}_2\text{O}_3 + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2\text{O} \]