Question

The reaction of 47.3mL of an aqueous HCl solution with excess MgCO3 produces 538mL of CO2 at 22.9 defrees Celsius and 0.940 atm. What's the molarity of the HCl solution?

Answer 1

Ans. Step 1: Given, Volume of CO­₂ produced = 538.0 mL = 0.538 L

            Temperature, T = 22.9⁰C = 296.05 K

            Pressure, P = 0.940 atm

# Ideal gas equation:           PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

# Putting the values in equation 1-

0.940 atm x 0.538 L = n x (0.0821 atm L mol^-1K^-1) x 296.05 K

            Or, n = 0.50572 atm L / 24.305705 atm L mol^-1

            Hence, n = 0.020807 mol

Therefore, moles of CO₂ during the reaction = 0.020807 mol

Step 2: Balanced reaction: MgCO₃(s) + 2HCl(aq) ----> MgCl₂(aq) + H₂O(l) + CO₂(g)

According to the stoichiometry of balanced reaction, 1 mol CO₂ is produced by 2 mol HCl.

So,

            Moles of HCl consumed = 2 x Moles of CO₂ produced

                                                            = 2 x 0.020807 mol

                                                            = 0.041614 mol

# Given, volume of HCl solution = 47.30 mL = 0.0473 L

Now,

            Molarity of HCl = Moles of HCl / Volume of solution in liters

                                    = 0.041614 mol / 0.0473 L

                                    = 0.879 mol/ L

                                    = 0.879 M

Hence, molarity of HCl solution = 0.879 M