Question

If 155 mL of wet H2 is collected over water at 24 ∘C and a barometric pressure of 739 torr , how many grams of Zn have been consumed? (The vapor pressure of water is 22.38 torr.)

Answer 1

The reaction between Zn and H2SO4 produces H2 gas.

         Zn(s) + H2SO4(aq) ----> ZnSO4 (aq) + H2 (g)

First we have to calculate the number of moles of H2 gas collected.

Given that 155 mL of wet H2 is collected over water at 24 ∘C and a barometric pressure of 739 torr.

             Hence, pressure of gas = barometric pressure - vapor presure of water

                                                   = 739 torr - 22.38 torr = 716.62 torr

                  pressure of gas, P = 716.62 torr = 716.62/760 atm = 0.943 atm

                 volume of H2, V = 155 mL = 0.155 L

                 temperature, T = 24^oC = 24 + 273 K = 297 K

    Ideal gas equation is PV = nRT

                            where R = universal gas constant = 0.0821 L.atm/K/mol

                                     n = number of moles of gas

                                       PV = n RT  

                                    n = PV/RT

                                       = [(0.943 atm) (0.155 L)]/ [(0.0821 L.atm/K/mol) (297K) ]

                                      = 0.006 moles

                                 n = 0.006 moles

      Therefore, 0.006 moles of H2 gas collected.

Calculation of how many grams of Zn have been consumed ?

        moles of H2 gs collected = 0.006

      Zn(s)   +     H2SO4(aq) ----> ZnSO4 (aq) + H2 (g)

     1 mol                                                             1 mol

    65.3 g                                                           1 mol              ( Molar mass of Zn = 65.3 g/mol)

      ? g                                                                0.006 mol

From this , we can say that

         To collect 1 mol of H2 gas , 65.3 g of Zn consumed.

So,    To collect 0.006 mol of H2 gas ,    ?   g of Zn consumed.

                   ? = (0.006 mol of H2 / 1 mol of H2) x 65.3 g of Zn

                      = 0.39 g of Zn

Mass of Zn consumed = 0.39 g

Therefore, 0.39 grams of Zn have been consumed.