Question

Calculate the volume of hydrogen gas liberated at stp when 32.65 gram of zinc reacts with hcl

Answer 1

Hello Dear.

Here is the answer---

Mass of the Zinc. = 32.65 grams.
Molar Mass of the Zinc. = 65.3 g/mole.
Molar Mass of Hydrogen = 1 g/mole.

Reaction when the Zinc reacts with HCl is →

Zn + 2HCl ------→ ZnCl₂ + H₂

From the Reaction,

∵ 65.3 g of Zn on reaction forming 1 × 2 g of Hydrogen.
∴ 1 g of Zn on reaction is forming 2/65.3 g of Hydrogen.
∴ 32.65 g on reaction is forming 1 g of Hydrogen.

Now,
We know, 1 mole of Hydrogen Occupies the Volume of 22.4 litres at S.T.P.
∴ 1 grams of the Hydrogen occupies the Volume of 22.4 litres at S.T.P.


Hope it helps.

Answer 2

Chemical equation:

Zn(s) + 2HCl (aq)-----> ZnCl2(aq) + H2(g)

Molecular mass of Zinc= 65.3u
Molar mass of Zn=65.3g

Molecular mass of H2=2u
Molar mass of H2=2g

According to equation:
Zn(s) + 2HCl (aq)-----> ZnCl2(aq) + H2(g)

1mole of Zn produces 1 mole of Hydrogen gas.

so,
1mole(65.3 g of Zn) on reacting with HCl-----> 1 mole of H2gas( 2g)

65.3g of Zn -----> 2g of H2
Now,
32.65g of Zn produces --> ?
mass of H2 gas evolved= 32.65x2/65.3
=65.3/65.3
=1g of Hydrogen gas

Hence 32.65 g of Zn reacts with HCl to produce 1 gramof Hydrogen gas.

As we know that , 1 mole of any gas at STP occupies 22.4 litres of gas.

2g of H2 occupies 22.4 litres at STP.
1 g of H2 occupies ---->?
So 1g of Hydrogen occupies 22.4/2 =11.2 litres

Hence 11.2 litres of Hydrogen gas is liberated when 32.65g of reacts with HCl.