10.Calculate the volume of oxygen that reacts with 500cm3 of hydrogen sulfide with the volume of both gases measured at rtp: 2H2S + 3O2 > 2SO2 + 2H20

0.79dm³

Explanation:

The reaction expression is given as:

            2H₂S   +   3O₂  →   2SO₂   +   2H₂O

Volume of hydrogen sulfide  = 500cm³ = 0.5dm³  

Unknown:

volume of oxygen that reacted  = ?

Solution:

 At rtp;        

                         22.4dm³ of gas has a mole of 1

            0.5dm³ will       = 0.022mole  of hydrogen sulfide;

                        2 mole of H₂S combines with 3 mole of O₂

           0.022 mole of H₂S will combine with   = 0.033mole of O₂

So;

   Since;

           1 mole of a gas occupied a volume of 22.4dm³ at rtp;

        0.033mole of Oxygen gas will occupy 0.033 x 22.4  = 0.79dm³

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%