Question

Quєѕtíσn :

Sample of impure magnesium is reacted with dilute sulphuric acid to give the respective salt and hydrogen. If 1 g. of the impure sample gave 298.6 cc. of hydrogen at s.t.p.

Calculate the % purity of the sample. [Mg = 24, H = 1].​

Answer 1

Explanation:

Refer to the attachment!!!!!!!!!!!!

Answer 2

Explanation:

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)