Question

According to the following reaction: 2 C6H6(g) + 15 O2(g) = 12 CO2(g) + 6 H2O(g)If 2.45 L of benzene are consumed in this reaction, how many liters of water vapor can be formed, if the reaction occurs at 1.25 ATM and 25 C?

Answer 1

Volume of water = 7.35 L

Step-by-step explanation

Given:

`2C_6H_6\text{ + 15O}_2\rightarrow12CO_2+\text{ 6H}_2O`

Volume of benzene = 2.45 L

Pressure = 1.25 atm

Temperature = 25 C = 298 K

Required: Volume of water vapour that will be formed

Solution

Step 1: Calculate the number of moles of benzene using ideal gas law

PV = nRT

n = PV/RT

n = (1.25 atm x 2.45 L)/(0.08206 L.atm/K.mol x 298 K)

n = 0.125 mol

Step 2: Use the stoichiometry to find the moles of water

The molar ratio between benzene and water is 2:6

Therefore moles of water = 0.125 mol x (6/2) =0.376 mol

Step 3: Calculate the volume of water that will be formed

V = nRT/P

V = (0.376 mol x 0.08206 L.atm/K.mol x 298 K)/1.25 atm

V = 7.35 L