To solve the problem, we need to determine how many liters of [tex]\( \text{C}_2\text{H}_2 \)[/tex] will react with [tex]\( 12.0 \)[/tex] moles of [tex]\( \text{O}_2 \)[/tex] at STP (Standard Temperature and Pressure). We will go through the following steps:
1. Determine the molar ratio of [tex]\( \text{C}_2\text{H}_2 \)[/tex] to [tex]\( \text{O}_2 \)[/tex] from the balanced equation:
The balanced chemical equation 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]
From this equation, we observe the molar ratio of [tex]\( \text{C}_2\text{H}_2 \)[/tex] to [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\frac{2 \text{ moles of } \text{C}_2\text{H}_2}{5 \text{ moles of } \text{O}_2}
\][/tex]
2. Calculate the moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] required for [tex]\( 12.0 \)[/tex] moles of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\text{moles of } \text{C}_2\text{H}_2 = 12.0 \text{ mol O}_2 \times \frac{2 \text{ mol } \text{C}_2\text{H}_2}{5 \text{ mol } \text{O}_2} = 4.8 \text{ mol } \text{C}_2\text{H}_2
\][/tex]
3. Calculate the volume of [tex]\( \text{C}_2\text{H}_2 \)[/tex] at STP:
At STP, one mole of any gas occupies [tex]\( 22.4 \)[/tex] liters. Therefore, the volume of [tex]\( 4.8 \)[/tex] moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] can be calculated as:
[tex]\[
\text{Volume of } \text{C}_2\text{H}_2 = 4.8 \text{ mol } \times 22.4 \text{ L/mol } = 107.52 \text{ L}
\][/tex]
Summary:
After the calculations, it is found that [tex]\( 4.8 \)[/tex] moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] are required, and their volume at STP is [tex]\( 107.52 \)[/tex] liters. Therefore, [tex]\( 12.0 \)[/tex] moles of [tex]\( \text{O}_2 \)[/tex] will react with [tex]\( 107.52 \)[/tex] liters of [tex]\( \text{C}_2\text{H}_2 \)[/tex] at STP.
