Let's solve this problem step-by-step using dimensional analysis as requested.
1. Step 1: Calculate the number of moles of [tex]\( O_2 \)[/tex] given: We know that at STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters.
[tex]\[
\text{Given volume of oxygen} = 13.5 \text{ L}
\][/tex]
We use the molar volume to convert this to moles of [tex]\( O_2 \)[/tex]:
[tex]\[
\text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{13.5 \text{ L}}{22.4 \text{ L/mol}} = 0.6026785714285715 \text{ mol}
\][/tex]
2. Step 2: Use the stoichiometric relationship to find the moles of [tex]\( H_2O \)[/tex] produced: The balanced chemical equation shows the stoichiometric relationship between [tex]\( O_2 \)[/tex] and [tex]\( H_2O \)[/tex]:
[tex]\[
2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g)
\][/tex]
From the equation, 5 moles of [tex]\( O_2 \)[/tex] produce 2 moles of [tex]\( H_2O \)[/tex]. Therefore, the ratio between [tex]\( O_2 \)[/tex] and [tex]\( H_2O \)[/tex] is:
[tex]\[
\frac{\text{Moles of } H_2O}{\text{Moles of } O_2} = \frac{2}{5}
\][/tex]
Now, using the moles of [tex]\( O_2 \)[/tex] calculated in Step 1:
[tex]\[
\text{Moles of } H_2O = \frac{2}{5} \times \text{Moles of } O_2 = \frac{2}{5} \times 0.6026785714285715 \approx 0.2410714285714286 \text{ mol}
\][/tex]
Thus, the number of moles of [tex]\( H_2O \)[/tex] formed from 13.5 liters of [tex]\( O_2 \)[/tex] at STP is approximately 0.2410714285714286 moles.
