Each of the following compounds decomposes to produce \(\mathrm{O}_{2}(\mathrm{g})\) when heated: \((\mathrm{a}) \mathrm{HgO}(\mathrm{s}) ;\) (b) \(\mathrm{KClO}_{4}(\mathrm{s})\) Write plausible equations for these reactions.

Short Answer

Expert verified
The balanced equations for the decomposition of HgO and KClO4 are: 2 HgO(s) → 2 Hg(l) + O2(g) and KClO4(s) → KCl(s) + 2 O2(g), respectively.

Step by step solution

01

Write Unbalanced Decomposition Reaction for HgO

In the decomposition of mercury(II) oxide (HgO), mercury and oxygen gas are formed. The reaction can be represented as: HgO(s) → Hg(l) + O2(g). But this is not balanced yet.
02

Balancing the Decomposition Reaction for HgO

The rule for balancing chemical equations is that the number of atoms for each element should be the same on both sides of the equation. There are 2 oxygen atoms on the right side while there is only 1 on the left side in our equation. To balance this, place a 2 in front of HgO, and 1 in front of Hg, which gives us the balanced equation is: 2 HgO(s) → 2 Hg(l) + O2(g)
03

Write Unbalanced Decomposition Reaction for KClO4

In the decomposition of potassium perchlorate (KClO4), potassium chloride and oxygen gas are formed. The reaction can be represented as: KClO4(s) → KCl(s) + O2(g). But this equation is not balanced yet.
04

Balancing the Decomposition Reaction for KClO4

Again, apply the rule that the number of atoms for each element in a chemical equation must be the same on both sides. In our equation, there are 4 oxygen atoms on the left side, but only 2 on the right side. To balance this, put a 2 in front of O2. This gives us our final balanced equation: KClO4(s) → KCl(s) + 2 O2(g).

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