Question

365g of hydrochloric acid reacts with 230g of sodium metal in a controlled environment. Calculate the number of moles and volume of hydrogen gas liberated at STP.​

Answer 1

Answer:

Standard Chemical Equation of this reaction:

• 2 HCl + 2 Na ⇒ 2 NaCl + H₂

That is 2 moles of HCl react with 2 moles of Na to give 1 mole of H₂.

According to the question,

365 g of HCl is reacted with 230 g of Sodium. Calculating the number of moles present in these mixtures we get:

$\implies \text{Number of moles} = \dfrac{\text{Given Mass}}{\text{Molar Mass}}\\\\\\\implies \text{Number of moles of HCl} = \dfrac{\text{365 g}}{\text{36.5 g}}\\\\\\\implies \text{Number of moles of HCl} = 10\:moles$

$\implies \text{Number of moles of Na} = \dfrac{\text{230 g}}{\text{23 g}}\\\\\\\implies \text{Number of moles of Na} = 10\;\:moles$

Therefore 10 moles of HCl is being reacted with 10 moles of Na. Hence the number of moles of H₂ would be:

⇒ n = 10/2 = 5 moles of H₂.

Hence 5 moles of H₂ is obtained after 10 moles of HCl reacts with 10 moles of Na.

We know that, 1 mole occupies 22.4 liters of volume at S.T.P. Hence for 5 moles it would be:

⇒ Volume = 5 × 22.4 = 112 liters.

Hence 5 moles of H₂ occupies 112 liters at S.T.P.

Answer 2

Given :-

365g of hydrochloric acid reacts with 230g of sodium metal in a controlled environment.

To Find :-

Calculate the number of moles and volume of hydrogen gas liberated at STP.​

Solution :-

We know that

Molar mass of hydrochloric acid ≈ 36.5 g/ml

Molar mass of sodium = 23 g/mol

Now

No. of moles of hydrochloric acid = 365/36.5

No. of moles of hydrochloric acid = 3650/365

No. of moles of hydrochloric acid = 10 moles

No. of moles of sodium = 230/23

No. of moles of sodium = 10 moles

No. of moles of hydrogen = 10/2

No. of moles of hydrogen = 5

1 moles = 22.4 l

5 moles = 5 × 22.4

5 moles = 5 × 224/10

5 moles = 112 l