Given the reaction:

[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]

2.00 mol of [tex]\(\text{C}_6\text{H}_6\)[/tex] reacts with oxygen according to the reaction above. What volume of water vapor, in liters, forms at STP?

[tex]\[ [?] \text{L} \text{H}_2\text{O} \][/tex]



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].