To solve this problem, we need to determine the volume of water vapor produced when 2.00 moles of benzene (C6H6) react according to the given chemical equation.
The balanced chemical equation is:
[tex]\[
2 \text{ C}_6\text{H}_6(g) + 15 \text{ O}_2(g) \rightarrow 12 \text{ CO}_2(g) + 6 \text{ H}_2\text{O}(g)
\][/tex]
Here is the step-by-step solution:
1. Determine the molar ratio between C6H6 and H2O:
- According to the balanced chemical equation, 2 moles of C6H6 produce 6 moles of H2O.
- Therefore, for every 1 mole of C6H6, 3 moles of H2O are produced.
[tex]\[
\frac{6 \text{ moles H}_2\text{O}}{2 \text{ moles C}_6\text{H}_6} = 3 \text{ moles H}_2\text{O per mole of C}_6\text{H}_6
\][/tex]
2. Calculate the moles of H2O produced:
- Given that there are 2.00 moles of C6H6:
[tex]\[
\text{Moles of H}_2\text{O} = 2.00 \text{ moles C}_6\text{H}_6 \times 3 \text{ moles H}_2\text{O per mole of C}_6\text{H}_6 = 6.00 \text{ moles H}_2\text{O}
\][/tex]
3. Convert moles of H2O to volume of H2O at STP:
- At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
[tex]\[
\text{Volume of H}_2\text{O} = 6.00 \text{ moles H}_2\text{O} \times 22.4 \text{ liters per mole} = 134.4 \text{ liters}
\][/tex]
Based on these calculations, the volume of water vapor produced is:
[tex]\[
134.4 \text{ liters of H}_2\text{O}
\][/tex]
