To balance the equation for the reaction between mercuric chloride (HgCl₂) and aluminum (Al) resulting in aluminum chloride (AlCl₃) and mercury (Hg), we need to ensure that the number of each type of atom is the same on both sides of the equation.
The unbalanced equation is as follows: \[ \text{HgCl}_2 + \text{Al} \rightarrow \text{AlCl}_3 + \text{Hg} \]
Balancing the Equation:
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Starting with the reactants:
- We have 2 chlorine atoms (from HgCl₂) and 1 aluminum atom.
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On the product side:
- AlCl₃ has 3 chlorine atoms and 1 aluminum atom, and there's also 1 mercury atom.
Balancing Steps:
- We first need to balance the chlorine atoms. Since there are 2 Cl in HgCl₂, we will need to adjust coefficients:
- We know that for every 1 HgCl₂, we require 1 Al, but since AlCl₃ requires 3 Cl, we need 3 times the number of HgCl₂.
So let's add coefficients in front of each compound. To balance chlorine, we need \( 3 \) AlCl₃, which means we need \( 3 \) HgCl₂:
\[ 3 \text{HgCl}_2 + 2 \text{Al} \rightarrow 2 \text{AlCl}_3 + 3 \text{Hg} \]
Final Balanced Equation:
- This leads to the balanced equation: \[ 3 \text{HgCl}_2 + 2 \text{Al} \rightarrow 2 \text{AlCl}_3 + 3 \text{Hg} \]
Coefficients:
- Coefficient in front of HgCl₂: 3
- Coefficient in front of Al: 2
- Coefficient in front of AlCl₃: 2
- Coefficient in front of Hg: 3
So the final balanced equation is: \[ 3 \text{HgCl}_2 + 2 \text{Al} \rightarrow 2 \text{AlCl}_3 + 3 \text{Hg} \]