2. A student performed two experiments to establish how effective manganese(IV) oxide was as a catalyst for the decomposition of hydrogen peroxide. The results below were obtained by carrying out these experiments with two different quantities of manganese(IV) oxide. The volume of the gas produced was recorded against time.

\begin{tabular}{|l|r|r|r|r|r|r|r|r|}
\hline Time/s & 0 & 30 & 60 & 90 & 120 & 150 & 180 & 210 \\
\hline \begin{tabular}{l}
Volume for \\
[tex]$0.3 \, \text{g/cm}^3$[/tex]
\end{tabular} & 0 & 29 & 55 & 79 & 98 & 118 & 133 & 146 \\
\hline \begin{tabular}{l}
Volume for \\
[tex]$0.5 \, \text{g/cm}^3$[/tex]
\end{tabular} & 0 & 45 & 84 & 118 & 145 & 162 & 174 & 182 \\
\hline
\end{tabular}

g. How long did it take for [tex]$60 \, \text{cm}^3$[/tex] of gas to be produced when the experiment was carried out using 0.5 g of the manganese(IV) oxide?

h. Write a balanced chemical equation for the decomposition of hydrogen peroxide.



Answer :

Let's break down the solution step by step for the given questions.

### Part (g)
How long did it take for [tex]\(60 \, \text{cm}^3\)[/tex] of gas to be produced when the experiment was carried out using [tex]\(0.5 \, \text{g}\)[/tex] of the manganese(IV) oxide?

1. Given Data:
- Time (s): [tex]\(0, 30, 60, 90, 120, 150, 180, 210\)[/tex]
- Volume for [tex]\(0.5 \, \text{g} / \text{cm}^3\)[/tex] manganese(IV) oxide (cm³): [tex]\(0, 45, 84, 118, 145, 162, 174, 182\)[/tex]

2. Determine the relevant time interval:
- The target volume is [tex]\(60 \, \text{cm}^3\)[/tex].
- From the given data, we can see [tex]\(45\)[/tex] cm³ at [tex]\(30\)[/tex] seconds and [tex]\(84\)[/tex] cm³ at [tex]\(60\)[/tex] seconds.

3. Interpolate to find the exact time:
- The volume [tex]\(60 \, \text{cm}^3\)[/tex] lies between [tex]\(45 \, \text{cm}^3\)[/tex] (at [tex]\(30 \, \text{s}\)[/tex]) and [tex]\(84 \, \text{cm}^3\)[/tex] (at [tex]\(60 \, \text{s}\)[/tex]).
- We can use linear interpolation to find the exact time [tex]\(t\)[/tex] it took to reach [tex]\(60 \, \text{cm}^3\)[/tex]:

[tex]\[ \text{Time taken} = 30 + \left(\frac{60 - 45}{84 - 45}\right) \times (60 - 30) \][/tex]

4. Calculate the interpolated time:

[tex]\[ \text{Time taken} = 30 + \left(\frac{15}{39}\right) \times 30 \][/tex]

[tex]\[ \text{Time taken} = 30 + \left(\frac{15 \times 30}{39}\right) \][/tex]

[tex]\[ \text{Time taken} = 30 + \frac{450}{39} \][/tex]

[tex]\[ \text{Time taken} \approx 30 + 11.538 \][/tex]

[tex]\[ \text{Time taken} = 41.538 \, \text{s} \][/tex]

So, it took approximately [tex]\(41.538\)[/tex] seconds for [tex]\(60 \, \text{cm}^3\)[/tex] of gas to be produced with [tex]\(0.5 \, \text{g}\)[/tex] of manganese(IV) oxide.

### Part (h)
Write a balanced chemical equation for the decomposition of hydrogen peroxide.

The balanced chemical equation for the decomposition of hydrogen peroxide is:

[tex]\[ 2 \, \text{H}_2\text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} + \text{O}_2 \][/tex]

In this reaction:
- [tex]\( \text{H}_2\text{O}_2 \)[/tex] represents hydrogen peroxide.
- [tex]\( \text{H}_2\text{O} \)[/tex] represents water.
- [tex]\( \text{O}_2 \)[/tex] represents oxygen gas.