To determine the balanced equation for the decomposition of barium chlorate [tex]\((Ba(ClO_3)_2)\)[/tex], we need to make sure that the number of atoms of each element is the same on both sides of the equation.
1. Identify Reactants and Products:
- Reactant: [tex]\( Ba(ClO_3)_2 \)[/tex]
- Products: [tex]\( BaCl_2 \)[/tex] and [tex]\( O_2 \)[/tex]
2. Write the Unbalanced Equation:
[tex]\[
Ba(ClO_3)_2 \rightarrow BaCl_2 + O_2
\][/tex]
3. Balance the Barium (Ba) Atoms:
- There is 1 barium atom on both sides already.
4. Balance the Chlorine (Cl) Atoms:
- There are 2 chlorine atoms in the barium chlorate and 2 chlorine atoms in barium chloride, so this is balanced.
5. Balance the Oxygen (O) Atoms:
- There are 6 oxygen atoms in [tex]\(Ba(ClO_3)_2\)[/tex] (since [tex]\(ClO_3\)[/tex] has 3 oxygens, and there are 2 [tex]\(ClO_3\)[/tex] groups).
- Each [tex]\(O_2\)[/tex] molecule contains 2 oxygen atoms.
To have all 6 oxygen atoms in [tex]\(O_2\)[/tex] molecules, we need 3 [tex]\(O_2\)[/tex] molecules:
[tex]\[
Ba(ClO_3)_2 \rightarrow BaCl_2 + 3 O_2
\][/tex]
However, both barium chloride and oxygen gas molecules need to be fully represented, and their coefficients must balance appropriately:
6. Adjust and Add Correct Coefficients:
- To balance the chlorine atoms, we need to have two [tex]\(Ba(ClO_3)_2\)[/tex] molecules decomposing.
[tex]\[
2 Ba(ClO_3)_2 \rightarrow 2 BaCl_2 + 6 O_2
- Now, we have 2 barium atoms, 4 chlorine atoms, and 12 oxygen atoms on the reactant side.
- On the product side, we have 2 barium atoms, 4 chlorine atoms, and 12 oxygen atoms (distributed as 6 \(O_2\)).
Thus, the balanced equation for the decomposition of barium chlorate is:
\[
2 Ba(ClO_3)_2(s) \rightarrow 2 BaCl_2(s) + 6 O_2(g)
\][/tex]
So, the correct answer among the given choices is:
[tex]\[
2 Ba(ClO_3)_2(s) \rightarrow 2 BaCl_2(s) + 6 O_2(g)
\][/tex]
