Answered

```latex
[tex]\[
2 C_2 H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2 O(g)
\][/tex]

How many moles of [tex]\( H_2 O \)[/tex] form from [tex]\( 13.5 \, L \, O_2 \)[/tex], assuming the reaction is at STP?

[tex]\[
\begin{array}{l}
\begin{array}{l|c|c}
13.5 \, L \, O_2 & 1 \, \text{mol} \, O_2 & 2 \, \text{mol} \, H_2 O \\
\hline
& 22.4 \, L \, O_2 & 5 \, \text{mol} \, O_2
\end{array} \\
\end{array}
\][/tex]

[tex]\[ \text{[?] mol} \, H_2 O \][/tex]
```



Answer :

GinnyAnswer
To determine how many moles of [tex]\(H_2O\)[/tex] form from 13.5 L of [tex]\(O_2\)[/tex] assuming the reaction occurs at Standard Temperature and Pressure (STP), we can follow these steps:

### Step 1: Identify the key components and relationships in the problem

1. Volume of [tex]\(O_2\)[/tex]: Given as 13.5 liters.
2. Molar volume of a gas at STP: 22.4 liters for 1 mole of gas.
3. Stoichiometry of the reaction:
- From the balanced chemical equation:
[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]
- The molar ratio of [tex]\(O_2\)[/tex] to [tex]\(H_2O\)[/tex] is 5:2.

### Step 2: Calculate the moles of [tex]\(O_2\)[/tex]

To find the number of moles of [tex]\(O_2\)[/tex] in 13.5 liters, we use the molar volume of a gas at STP.

[tex]\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{13.5 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.6027 \, \text{mol} \][/tex]

### Step 3: Determine the moles of [tex]\(H_2O\)[/tex] produced

Using the stoichiometric relationship from the balanced equation, we convert moles of [tex]\(O_2\)[/tex] to moles of [tex]\(H_2O\)[/tex]. According to the equation, 5 moles of [tex]\(O_2\)[/tex] produce 2 moles of [tex]\(H_2O\)[/tex].

[tex]\[ \text{Moles of } H_2O = \text{Moles of } O_2 \times \left( \frac{2 \, \text{mol } H_2O}{5 \, \text{mol } O_2} \right) = 0.6027 \, \text{mol} \times \left( \frac{2}{5} \right) \approx 0.2411 \, \text{mol} \][/tex]

### Conclusion

Upon completing the calculations, the number of moles of [tex]\(H_2O\)[/tex] produced from 13.5 liters of [tex]\(O_2\)[/tex] is approximately 0.2411 moles.

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