The balanced reaction is shown below.
                       Â
Zn   +   Â
2HCl     ……………..> Â
ZnCl2 Â Â Â Â +
      H2
Mass: Â Â Â Â Â Â Â 65.38
g      2 x 36.5
g                Â
136.29 g       2.016 g
65.38 g of Zn produces H2 = 2.016 g
0.53 g of Zn produces H2 = (2.016 x 0.53)/65.38 g = 0.01634
g
Mole of H2 produced = 0.01634/2.016 mol = 0.0081 mol
We have the relation PV = nRT
Here, P = 1 atm, n = 0.0081 mol, R = 0.082 L.atm/K.mol, T = 273
K (considering STP condition)
Using the above relation V = nRT/P = (0.0081 x 0.082 x 273)/1 =
0.181 L
Volume of H2 gas generated is 0.181 L.
