sample of impure Magnesium is reacted with diluted SulphuricAcid to give the respective salt and Hydrogen.If 1 gram of impure samplegave 298.6 cc of Hydrogen at STP.Calculate the % purity of the sample(MAgnesium = 24,Hydrogen = 1)

Magnesium is reacted with diluted Sulphuric Acid gives the following reaction:

         Moles of H₂ at STP = (298.6)/(22.4 x 1000) = 0.0133  

         Mg        +         H₂SO₄  ------->   MgSO₄            +              H₂

                                                                                                      0.0133

Thus, stoich. coeff. are equal hence moles of Mg are same as H₂ .i.e = 0.0133

Now, Mg in gm = moles x molar mass of Mg

                           = 0.0133 x 24

                           =0.3192 gm

Now percentage of purity of the sample is = 0.3192/1 x 100

                                                                            = 31.9 or 32 % (approx)

Plz Approve!

Plz Approve!

Plz Approve!