Whether or not the reaction above can occur and result in an actual chemical change, we can write and balance an equation. Use the drop downs to add the correct coefficients to balance the equation. (4 points)

Question

HgCl2
+
Al
→
AlCl3
+
Hg
Numbers are 1, 2, 3, 4

Answer 1

To balance the chemical equation:

Unbalanced reaction: \[ \text{HgCl}_2 + \text{Al} \rightarrow \text{AlCl}_3 + \text{Hg} \]

Step 1: Count the atoms of each element on both sides.

• Reactants:

  • Hg: 1
  • Cl: 2
  • Al: 1

• Products:

  • Al: 1
  • Cl: 3
  • Hg: 1

Step 2: Balance the chlorine atoms.

To balance the Cl atoms, we need to change the coefficient of \(\text{AlCl}_3\) to 2:

Updated equation: \[ \text{HgCl}_2 + \text{Al} \rightarrow 2\text{AlCl}_3 + \text{Hg} \]

This gives us:

• Reactants:

  • Hg: 1
  • Cl: 2
  • Al: 1

• Products:

  • Al: 2
  • Cl: 6 (2 x 3 from \(\text{AlCl}_3\))
  • Hg: 1

Step 3: Balance the aluminum and mercury atoms.

Now, let’s balance the aluminum by placing a coefficient of 2 in front of Al in the reactants:

Balanced equation: \[ \text{HgCl}_2 + 2\text{Al} \rightarrow 2\text{AlCl}_3 + \text{Hg} \]

Final counts:

• Reactants:

  • Hg: 1
  • Cl: 2
  • Al: 2

• Products:

  • Al: 2
  • Cl: 6
  • Hg: 1

Now we can balance the Cl by putting a coefficient of 3 before \(\text{Hg}\):

Final balanced equation: \[ \text{HgCl}_2 + 3\text{Al} \rightarrow 2\text{AlCl}_3 + 3\text{Hg} \]

Thus, the balanced coefficients are:

• HgCl2: 1
• Al: 3
• AlCl3: 2
• Hg: 3

So, the final coefficients are: 1, 3, 2, 3