Question

0.05g of magnesium when treated with dilute HCl gave 51ml of hydrogen at 27°C and 780 mm pressure .calculate the equivalent mass of magnesium.

please solve it......​

Answer 1

Answer:

Explanation:

Mg+2HCl→MgCl2+H2

Mg+2H+

→

Mg2++H2

As 0.05g of Magnesium (Mg) is giving 51mL of Hydrogen gas on reaction with dil. HCl So, number of moles of hydrogen produced will determined by the formula PV=nRT

where, PressureP =780mmHg

Volume (V)=51mL=0.051L

Temperature (T)=270C=300K

R gasconstant =62.363mmHgL mol −1K−1

So,n=PV/RT

=780×0.051/62.363×300

=2.126×10−3moles

From the above calculation,mass of Mg reacted to produce 2.126×10−3 moles of Hydrogen =0.05g So, mass of Mg will react to produce 1 mole of Hydrogen =0.05/(2.126×10−3)=23.518g (i.e. Atomic mass of Mg) As Mg is displacing two moles of Hydrogen ions so its Equivalent mass will be = Atomic mass/2 =23.518/2=11.759g