asked by kartiksharma3228, 4 days ago

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

78
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.
10

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.

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