In a reaction of Zn (0.53 g) & HCl what is the volume of Hydrogen gas...

In a reaction of Zn (0.53 g) & HCl what is the volume of Hydrogen gas generated? (assume atmospheric pressure)

Expert Solution

The balanced reaction is shown below.

Zn + 2HCl ……………..> ZnCl₂ + H₂

Mass: 65.38 g 2 x 36.5 g 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.

queen_honey_blossom answered 2 years ago

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