To determine how many moles of [tex]\( H_2O \)[/tex] form from 13.5 L of [tex]\( O_2 \)[/tex] at Standard Temperature and Pressure (STP), we can follow this step-by-step procedure:
1. Determine the Moles of [tex]\( O_2 \)[/tex]:
- At STP, 1 mole of an ideal gas occupies 22.4 liters.
- Given volume of [tex]\( O_2 \)[/tex] = 13.5 L.
- We use the formula:
[tex]\[
\text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}}
\][/tex]
Substituting the given values:
[tex]\[
\text{Moles of } O_2 = \frac{13.5 \text{ L}}{22.4 \text{ L/mol}} \approx 0.6027 \text{ moles}
\][/tex]
2. Use the Stoichiometric Ratio to Find Moles of [tex]\( H_2O \)[/tex]:
- According to the balanced chemical equation:
[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 number of moles of [tex]\( H_2O \)[/tex] produced can be calculated using the proportionality:
[tex]\[
\text{Moles of } H_2O = \left( \frac{2 \text{ moles } H_2O}{5 \text{ moles } O_2} \right) \times \text{Moles of } O_2
\][/tex]
Substituting the known value:
[tex]\[
\text{Moles of } H_2O = \left( \frac{2}{5} \right) \times 0.6027 \approx 0.2411 \text{ moles}
\][/tex]
Thus, the number of moles of [tex]\( H_2O \)[/tex] formed from 13.5 L of [tex]\( O_2 \)[/tex] at STP is approximately 0.2411 moles.
