Question

10 ml of liquid carbon disulfide( specific gravity- 2.63) is burnt in oxygen.. find the volume of the gas at stp

Answer 1

Hello Dear.

Reaction of Carbon Disulphide with the Oxygen is ⇒

CS₂ + 3O₂ ----------------→ CO₂ + 2SO₂

Specific Gravity = 2.63

∵ Specific Gravity = Density of CS₂ ÷ Density of Water.
∴ 2.63 × 1 = Density of CS₂.
⇒ Density of CS₂ = 2.63 g/ml.

Now, Density = Mass/Volume.
∴ Mass of CS₂ = Density × Volume.
⇒ Mass of CS₂ = 2.63 × 10
∴ Mass of CS₂ = 26.3 g.

Now, 
Molar Mass of CS₂ = 76 g/mole.

We know, 1 mole of Gas at S.T.P. Occupies the volume of the 22.4 liter.

∴ From the Reaction,
Volume of CO₂ = 22.4 liter.
Volume of SO₂ = 22.4 × 2
= 44.8 liter.

Total Volume occupied by the mixture of Carbon dioxide and Sulphur dioxide is 22.4 + 44.8 = 67.2 liter.

Now, From the Reaction,

∵ 76 g of CS₂ Produces the 67.2 liter of Gases at S.T.P.
∴ 1 g of CS₂ Produces 67.2/76 liter of Gases at S.T.P.
∴ 26.3 g of CS₂ Produces 23.25 liter of Gases at S.T.P.


Hence, the volume occupied by the Gases at S.T.P. is 23.25 liter.


Hope it helps.

Answer 2

Answer:

Reaction of Carbon Disulphide with the Oxygen is ⇒

CS₂ + 3O₂ ----------------→ CO₂ + 2SO₂

Specific Gravity = 2.63

∵ Specific Gravity = Density of CS₂ ÷ Density of Water.

∴ 2.63 × 1 = Density of CS₂.

⇒ Density of CS₂ = 2.63 g/ml.

Now, Density = Mass/Volume.

∴ Mass of CS₂ = Density × Volume.

⇒ Mass of CS₂ = 2.63 × 10

∴ Mass of CS₂ = 26.3 g.

Now, 

Molar Mass of CS₂ = 76 g/mole.

We know, 1 mole of Gas at S.T.P. Occupies the volume of the 22.4 liter.

∴ From the Reaction,

Volume of CO₂ = 22.4 liter.

Volume of SO₂ = 22.4 × 2

= 44.8 liter.

Total Volume occupied by the mixture of Carbon dioxide and Sulphur dioxide is 22.4 + 44.8 = 67.2 liter.

Now, From the Reaction,

∵ 76 g of CS₂ Produces the 67.2 liter of Gases at S.T.P.

∴ 1 g of CS₂ Produces 67.2/76 liter of Gases at S.T.P.

∴ 26.3 g of CS₂ Produces 23.25 liter of Gases at S.T.P.