LUSSAC'S LAW
Acetylene (C2H2) burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burns 50cm^3 of acetylene. (Assume air contains 20% oxygen.

Open in App

Solution

C₂H₂+ 5/2O₂----------------> 2CO₂+ H₂O

Burning of 1 mole of C₂H₂requires 2.5 mole O₂

That means 2.5 x 22.4 L = 56000 cm³O₂

Let total air required is = M

20% of air = 56000 cm³O₂

20 M/100 = 56000 cm³O₂

M = 56000 x 100/20 = 280000 cm³

Suggest Corrections

Similar questions

Q. Acetylene

[C_{2} H_{2}]

burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burn

50 c m^{3}

of acetylene (Assume air contains 20 % oxygen):

What volume fo oxygen would be required to burn completely 400 ml. of acetylene $[C_{2} H_{2}]$ Also, calculated the volume fo carbon dioxide formed.
$2 C_{2} H_{2} + 50_{2} \to 4 C O_{2} + 2 H_{2} O (l)$

Q. What volume of oxygen would be required to burn completely 200 ml of acetylene

(C_{2} H_{2})

, and what would be the volume of carbon dioxide formed?

Q. Air contains 20% oxygen by volume. Calculate the theoretical volume of air which will be required for burning

200 m^{3}

of acetylene gas completely. All volumes are measured under the same conditions of temperature and pressure

Q. What volume of oxygen would be required to burn completely 400 ml of acetylene(

C_{2} H_{2}

Join BYJU'S Learning Program

LUSSAC'S LAW Acetylene (C2H2) burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burns 50cm^3 of acetylene. (Assume air contains 20% oxygen.

C2H2 + 5/2O2 ----------------> 2CO2 + H2O Burning of 1 mole of C2H2 requires 2.5 mole O2 That means 2.5 x 22.4 L = 56000 cm3 O2 Let total air required is = M 20% of air = 56000 cm3 O2 20 M/100 = 56000 cm3 O2 M = 56000 x 100/20 = 280000 cm3

LUSSAC'S LAW
Acetylene (C2H2) burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burns 50cm^3 of acetylene. (Assume air contains 20% oxygen.

C₂H₂+ 5/2O₂----------------> 2CO₂+ H₂O

Burning of 1 mole of C₂H₂requires 2.5 mole O₂

That means 2.5 x 22.4 L = 56000 cm³O₂

Let total air required is = M

20% of air = 56000 cm³O₂

20 M/100 = 56000 cm³O₂

M = 56000 x 100/20 = 280000 cm³