Question

In the oxidation of hydrogen chloride gas with air to produce chlorine (4 HCl + O₂ -> 2 Cl₂ + 2 H₂O), where 50% excess oxygen is used and the reaction is only 80% complete, calculate the following:

Determine the volume of air admitted per 100 m³ of HCl if both air and HCl enter the reactor at 1 bar and 290 K.

Answer 1

To determine the volume of air admitted per 100 m³ of HCl, use the ideal gas law to calculate the moles of HCl and the moles of Cl₂ produced. Then, use the stoichiometry of the balanced equation to calculate the volume of air. The volume of air admitted per 100 m³ of HCl is 8054.2 L.

Step-by-step explanation:

In the oxidation of hydrogen chloride gas with air to produce chlorine (4 HCl + O₂ -> 2 Cl₂ + 2 H₂O), the volume of air admitted per 100 m³ of HCl can be determined using the ideal gas law.

First, calculate the moles of HCl using the given volume and conditions:

n(HCl) = PV / RT

= (1 bar)(100 m³)(1 L / 1000 mL)(101.3 kPa / 1 bar) / (0.0821 L•atm / K•mol)(290 K)

= 910.99 mol

Next, calculate the moles of Cl₂ produced, considering that the reaction is only 80% complete:

n(Cl₂) = 80% of n(HCl)

= 0.8 * 910.99 mol

= 728.79 mol

Finally, calculate the volume of air admitted per 100 m³ of HCl using the stoichiometry of the balanced equation:

Volume of air = (2 * n(Cl₂) * 22.4 L) / (4 mol)

= (2 * 728.79 mol * 22.4 L) / (4)

= 8054.2 L