Question

the decomposition of hydrogen peroxide in the presence of a catalyst at standard pressure and room temperature is given by the unbalanced chemical equation below.
H_2O_2 (aq) - >H_2O (litre) +O_2(g)
Take the molar gas volume (Vm) as 24,45 dm^3 at room temperature and standard temperature.
balanced the equation

Answer 1

To balance the decomposition reaction of hydrogen peroxide (H₂O₂) into water (H₂O) and oxygen gas (O₂), we start with the unbalanced equation:

\[
\text{H}_2\text{O}_2(aq) \rightarrow \text{H}_2\text{O}(l) + \text{O}_2(g)
\]

Now, let's balance the equation step by step:

1. **Count the atoms for each element** on both sides of the reaction:
- Reactants: 2 H and 2 O (from H₂O₂)
- Products: 2 H (from H₂O) and 2 O (1 from H₂O and 2 from O₂)

2. **Balance the equation** by adjusting the coefficients:
- We have 2 oxygen atoms in H₂O₂ and 1 oxygen atom in H₂O and 2 oxygen atoms in O₂ on the product side. To balance the oxygen atoms, we note that 2 H₂O₂ will produce 2 H₂O and 1 O₂.

So the balanced equation becomes:

\[
2 \text{H}_2\text{O}_2(aq) \rightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g)
\]

In this balanced equation:
- There are 4 hydrogen atoms on both sides (2 H₂O₂ → 4 H in products).
- There are 4 oxygen atoms on both sides (2 O from 2 H₂O and 2 O in O₂).

This equation is now balanced.