Question

When 2 L of NH3 (g) and 1.2 L of HCl (g) are mixed then volume of gaseous mixture after the reaction will be {NH3(g) + HCl(g) → NH4Cl(s)}
A)1.2 L
B)2 L
C)0.8 L
D)3.2 L

Answer 1

Answer:

Explanation:

1mole = 22.4 L

Thus miles of NH3 = 2/22.4

Moles of HCl = 1.2/22.4

Clearly HCl is the limiting reagent.

As per the reaction one mole of HCl gives one mole of nh4cl. So,

1.2/22.4 moles of HCl will give 1.2/22.4 moles of nh4cl.

Now,

1mole = 22.4 L

1.2/22.4 moles will be equal to 1.2 L.

Thus volume after the reaction occurs is 1.2 L.

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Answer 2

Given:

Volume of NH3 = 2 L

Volume of HCl = 1.2 L

To Find:

The volume of the gaseous mixture after the reaction.

Calculation:

- Here it is clear that HCl is the limiting reagent.

- According to the given reaction:

22.4 L of HCl reacts with 22.4 L of NH3 to give 22.4 L of NH4Cl.

⇒ 1.2 L of HCl will react with 1.2 L of NH3 to give 1.2 L of NH4Cl.

⇒ The volume of NH4Cl = 1.2 L

& the volume of NH3 left unreacted = 2 - 1.2 = 0.8 L

- The total volume of gaseous mixture after the reaction = 1.2 + 0.8

⇒ V = 2 L

- So, the correct answer is option (B) 2 L.