Certainly! Let's solve this step-by-step.
1. Identify the Given Reaction and Known Quantities:
The balanced chemical equation provided is:
[tex]\[
2 \, C_6H_6(g) + 15 \, O_2(g) \rightarrow 12 \, CO_2(g) + 6 \, H_2O(g)
\][/tex]
We are given that 2.00 moles of [tex]\(C_6H_6\)[/tex] reacts with oxygen.
2. Identify the Relevant Stoichiometric Ratios:
According to the balanced chemical equation, 2 moles of [tex]\(C_6H_6\)[/tex] produce 6 moles of [tex]\(H_2O\)[/tex]. This can be written as a stoichiometric ratio:
[tex]\[
\frac{6 \, \text{mol} \, H_2O}{2 \, \text{mol} \, C_6H_6}
\][/tex]
3. Calculate the Moles of [tex]\(H_2O\)[/tex] Produced:
Since we start with 2.00 moles of [tex]\(C_6H_6\)[/tex]:
[tex]\[
\text{Moles of } H_2O = \left( \frac{6 \, \text{mol} \, H_2O}{2 \, \text{mol} \, C_6H_6} \right) \times 2.00 \, \text{mol} \, C_6H_6
\][/tex]
Simplifying the expression, we get:
[tex]\[
\text{Moles of } H_2O = 6.00 \, \text{mol}
\][/tex]
4. Convert Moles of [tex]\(H_2O\)[/tex] to Volume at STP:
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, we can find the volume of water vapor produced using the number of moles calculated:
[tex]\[
\text{Volume of } H_2O = 6.00 \, \text{mol} \times 22.4 \, \text{L/mol}
\][/tex]
Simplifying this, we obtain:
[tex]\[
\text{Volume of } H_2O = 134.4 \, \text{L}
\][/tex]
Thus, the volume of water vapor produced at STP is [tex]\(134.4 \, \text{liters}\)[/tex].
