To determine the number of moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced from 13.5 liters of [tex]\( \text{O}_2 \)[/tex] at STP, let's follow the steps:
1. Identify the reaction and relevant stoichiometry:
The balanced chemical equation for the combustion of acetylene ([tex]\( \text{C}_2\text{H}_2 \)[/tex]) is:
[tex]\[
2 \text{C}_2\text{H}_2(g) + 5 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 2 \text{H}_2\text{O(g)}
\][/tex]
According to this balanced equation, 5 moles of [tex]\( \text{O}_2 \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
2. Calculate the moles of [tex]\( \text{O}_2 \)[/tex] present:
Given a volume of [tex]\( 13.5 \)[/tex] liters of [tex]\( \text{O}_2 \)[/tex] at STP (Standard Temperature and Pressure), use the molar volume of a gas at STP, which is [tex]\( 22.4 \)[/tex] liters per mole, to find the number of moles of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\text{moles of } \text{O}_2 = \frac{13.5 \text{ L}}{22.4 \text{ L/mol}} \approx 0.6027 \text{ moles}
\][/tex]
3. Use stoichiometry to find moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced:
From the balanced equation, we know that 5 moles of [tex]\( \text{O}_2 \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore, the ratio of moles of [tex]\( \text{H}_2\text{O} \)[/tex] to moles of [tex]\( \text{O}_2 \)[/tex] is:
[tex]\[
\text{moles of } \text{H}_2\text{O} = \frac{2}{5} \times \text{moles of } \text{O}_2
\][/tex]
Plugging in the number of moles of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\text{moles of } \text{H}_2\text{O} = \frac{2}{5} \times 0.6027 \approx 0.2411 \text{ moles}
\][/tex]
So, from 13.5 liters of [tex]\( \text{O}_2 \)[/tex] at STP, approximately [tex]\( 0.6027 \)[/tex] moles of [tex]\( \text{O}_2 \)[/tex] and [tex]\( 0.2411 \)[/tex] moles of [tex]\( \text{H}_2\text{O} \)[/tex] are produced.
