Answer :
Alright, let's solve this step by step.
1. Analyze the Balanced Chemical Equation:
The balanced chemical equation given 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]
From this equation, we observe that 2 moles of benzene (C₆H₆) produce 6 moles of water (H₂O).
2. Determine the Moles of Water Produced:
Given that we start with 2.00 moles of C₆H₆, we can use the mole ratio from the balanced equation to find the moles of water produced. The ratio is:
[tex]\[ \frac{6 \text{ moles of H}_2\text{O}}{2 \text{ moles of C}_6\text{H}_6} = 3 \][/tex]
Therefore, the moles of H₂O produced from 2.00 moles of C₆H₆ are:
[tex]\[ 2.00 \text{ moles C}_6\text{H}_6 \times 3 = 6.00 \text{ moles H}_2\text{O} \][/tex]
3. Calculate the Volume of Water Vapor at STP:
At Standard Temperature and Pressure (STP), one mole of any ideal gas occupies 22.4 liters. Therefore, we can find the volume of water vapor by multiplying the moles of H₂O by 22.4 liters per mole.
[tex]\[ \text{Volume of H}_2\text{O} \text{ at STP} = 6.00 \text{ moles} \times 22.4 \text{ L/mol} \][/tex]
Simplifying this, we get:
[tex]\[ \text{Volume of H}_2\text{O} \text{ at STP} = 134.4 \text{ liters} \][/tex]
So, the volume of water vapor formed at STP is [tex]\(134.4 \text{ liters}\)[/tex].
1. Analyze the Balanced Chemical Equation:
The balanced chemical equation given 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]
From this equation, we observe that 2 moles of benzene (C₆H₆) produce 6 moles of water (H₂O).
2. Determine the Moles of Water Produced:
Given that we start with 2.00 moles of C₆H₆, we can use the mole ratio from the balanced equation to find the moles of water produced. The ratio is:
[tex]\[ \frac{6 \text{ moles of H}_2\text{O}}{2 \text{ moles of C}_6\text{H}_6} = 3 \][/tex]
Therefore, the moles of H₂O produced from 2.00 moles of C₆H₆ are:
[tex]\[ 2.00 \text{ moles C}_6\text{H}_6 \times 3 = 6.00 \text{ moles H}_2\text{O} \][/tex]
3. Calculate the Volume of Water Vapor at STP:
At Standard Temperature and Pressure (STP), one mole of any ideal gas occupies 22.4 liters. Therefore, we can find the volume of water vapor by multiplying the moles of H₂O by 22.4 liters per mole.
[tex]\[ \text{Volume of H}_2\text{O} \text{ at STP} = 6.00 \text{ moles} \times 22.4 \text{ L/mol} \][/tex]
Simplifying this, we get:
[tex]\[ \text{Volume of H}_2\text{O} \text{ at STP} = 134.4 \text{ liters} \][/tex]
So, the volume of water vapor formed at STP is [tex]\(134.4 \text{ liters}\)[/tex].
