Question

6.5g of zinc was reacted with excess of dilute hcl calculate the amount and volume of hydrogen produced at STP
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Answer 1

Answer ⇒ Amount of Hydrogen = 0.2 g.

Volume of hydrogen = 2.24 L.

Explanation ⇒ Reaction of Zinc with HCl.

Zn + 2HCl -------⇒ ZnCl₂ + H₂

From this reaction,

1 mole of Zn reacts with 2 moles of HCl.

∴ 65 g of Zn reacts with 2 × 36.5 g of HCl.

∴ 65 g of Zn reacts with 73 g of HCl.

∴  1 g of Zn reacts with 73/65 g of HCl.

∴ 6.5 g of Zn reacts with 73/65 × 6.5 g = 7.3 g of HCl.

Now, Zn is in less amount. Thus it is the limiting reagent which will govern the reaction.

∴ 1 mole of Zn produces 1 mole of Hydrogen gas.

∴ 65 g of Zn produces 2 g of Hydrogen.

∴ 6.5 g of Zn produces 2/65 × 6.5 g = 0.2 g of hydrogen.

Now, 2 g of Hydrogen occupies an volume of 22.4 L.

∴ Volume of 0.2 g of hydrogen = 22.4/2 × 0.2

= 2.24 L.

∴ Amount of Hydrogen = 0.2 g.

Volume of hydrogen = 2.24 L.

Hope it helps.

Answer 2

Answer:

Amount of hydrogen produced at STP is 2g

Volume of Hydrogen produced is 2.24L

Explanation:

Given Zinc reacts with Hydro chloric acid to produce zinc chloride and hydrogen gas.

Balanced chemical equation is :

Zn +2HCl --->ZnCl2 +H2

We can say from above equation that :

1 mole of Zinc reacts reacts with 2 moles of hydro chloric acid to produce 1 ole of Zinc chloride and 1 mole of Hydrogen gas.

According to question : 6.5 g of Zn reacts with excess of HCl.

so excess reactant is HCl

Limiting reactant = Zn

hence the production of products depend upon Zn

so, According to balanced equation :

65g of Zn produces  2g of H2

6.5 g of Zn produces how much ?

65g of Zn--------------->2g of H2

6.5g of Zn-------------->?

= 2x6.5/65

=0.2g

So amount of H2 produced is 0.2g

1mole of any gas occupies 22.4 L at STP

so 1 mole [ 2g] of H2 occupies 22.4 L

0.2 g of H2 occupies --------->?

=22.4x0.2/2=2.24L

so 2.24L of H2 gas is produced.