According to the following reaction: ___ C6H6(g) + ___ O2(g)→___ CO2(g) + ___ H2O(g)a. If 2.45 liters of benzene are consumed in this reaction, how many liters of water vapor can be formed, if thereaction occurs at 1.25 atm and 25°C?



Answer :

Answer:

7.33L of water vapor can be formed.

Explanation:

1st) It is necessary to balance the chemical reaction:

[tex]2C_6H_6+15O_2\rightarrow12CO_2+6H_2O[/tex]

From the balanced reaction we know that with 2 moles of benzene (C6H6), 6 moles of water vapor can be formed.

2nd) We need to convert the 2.45 liters of benzene to moles, using the Gas Law formula:

[tex]\begin{gathered} P*V=n*R*T \\ 1.25atm*2.45L=n*0.082\frac{atm*L}{mol*K}*298K \\ 3.06atm*L=n*24.44\frac{atm*L}{mol} \\ \frac{3.06atm*L}{24.44\frac{atm*L}{mol}}=n \\ 0.125mol=n \end{gathered}[/tex]

Now we know that 0.125 moles of benzene were consumed.

3rd) From the stoichiometry of the reaction and the 0.125 moles, we can calculate the moles of water vapor that can be formed:

[tex]\begin{gathered} 2molesC_6H_6-6molesH_2O \\ 0.125molesC_6H_6-x=\frac{0.125molesC_6H_6*6molesH_2O}{2molesC_6H_6} \\ x=0.375molesH_2O \end{gathered}[/tex]

0.375 moles of water vapor are formed.

4th) Finally, we can use the Gas Law formula again, to calculate the liters of water vapor that can be formed:

[tex]\begin{gathered} P*V=n*R*T \\ 1.25atm*V=0.375mol*0.082*\frac{atmL}{molK}*298K \\ 1.25atm*V=9.16atm*L \\ V=\frac{9.16atm*L}{1.25atm} \\ V=7.33L \end{gathered}[/tex]

So, 7.33L of water vapor can be formed.

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