1. Test - CHG101 (25. March 2023) 4. Selenium is an essential trace element. And while a healthy diet with lots of grains nonually provides enough of this chalcogen, supplements are available. One of them is Selenophoszald, $$ \mathrm{C}_{12} \mathrm{H}_{12} \mathrm{NPSe} $$. a) Set up a stoichiometric reaction equation for the combustion of Selenophoszald! Assume that selenium, Se , reaches its highest valency! [ 6 CP$$ ] $$

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To set up a stoichiometric reaction equation for the combustion of Selenophoszald ($$ \mathrm{C}_{12} \mathrm{H}_{12} \mathrm{NPSe} $$), we need to consider the products of combustion, which typically include carbon dioxide ($$ \mathrm{CO}_2 $$), water ($$ \mathrm{H}_2\mathrm{O} $$), and in this case, selenium will likely form selenium dioxide ($$ \mathrm{SeO}_2 $$) when it reaches its highest valency.

### Step 1: Write the unbalanced combustion reaction

The general form of the combustion reaction for an organic compound is:

$$
\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{N}_{z} \mathrm{P}_{a} \mathrm{Se}_{b} + O_2 \rightarrow CO_2 + H_2O + N_2 + P_2O_5 + SeO_2
$$

For Selenophoszald, we have:

$$
\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{NPSe} + O_2 \rightarrow CO_2 + H_2O + N_2 + SeO_2
$$

### Step 2: Identify the number of each element

- **Carbon (C)**: 12
- **Hydrogen (H)**: 12
- **Nitrogen (N)**: 1
- **Selenium (Se)**: 1
- **Oxygen (O)**: To be determined from the products.

### Step 3: Write the products

1. **Carbon**: 12 carbon atoms will produce 12 $$ \mathrm{CO}_2 $$.
2. **Hydrogen**: 12 hydrogen atoms will produce 6 $$ \mathrm{H}_2\mathrm{O} $$ (since each water molecule contains 2 hydrogen atoms).
3. **Nitrogen**: 1 nitrogen atom will produce $$ \frac{1}{2} \mathrm{N}_2 $$ (since nitrogen gas is diatomic).
4. **Selenium**: 1 selenium atom will produce 1 $$ \mathrm{SeO}_2 $$.

### Step 4: Write the balanced equation

Now we can write the balanced equation:

$$
\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{NPSe} + O_2 \rightarrow 12 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O} + \frac{1}{2} \mathrm{N}_2 + \mathrm{SeO}_2
$$

### Step 5: Balance the oxygen atoms

Now we need to balance the oxygen atoms. The products contain:

- From $$ 12 \mathrm{CO}_2 $$: $$ 12 \times 2 = 24 $$ oxygen atoms
- From $$ 6 \mathrm{H}_2\mathrm{O} $$: $$ 6 \times 1 = 6 $$ oxygen atoms
- From $$ \mathrm{SeO}_2 $$: $$ 1 \times 2 = 2 $$ oxygen atoms

Total oxygen needed:

$$
24 + 6 + 2 = 32 \text{ oxygen atoms}
$$

Since $$ O_2 $$ provides 2 oxygen atoms per molecule, we need:

$$
\frac{32}{2} = 16 \text{ molecules of } O_2
$$

### Final balanced equation

Putting it all together, the balanced combustion reaction for Selenophoszald is:

$$
\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{NPSe} + 16 \mathrm{O}_2 \rightarrow 12 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O} + \frac{1}{2} \mathrm{N}_2 + \mathrm{SeO}_2
$$

This is the stoichiometric reaction equation for the combustion of Selenophoszald.
The stoichiometric reaction for the combustion of Selenophoszald ($$ \mathrm{C}_{12} \mathrm{H}_{12} \mathrm{NPSe} $$) is: $$ \mathrm{C}_{12} \mathrm{H}_{12} \mathrm{NPSe} + 16 \mathrm{O}_2 \rightarrow 12 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O} + \frac{1}{2} \mathrm{N}_2 + \mathrm{SeO}_2 $$

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