200 ml of C 2 H 4 was burnt in sufficient air containing 20 - oxygen as per the equation C 2 H 4-3 O 2 - - - - - - - - 2 CO 2-2 H 2 O- Calculate the resultant mixture composition at 100- C and constant pressure-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Volume of Gases and Number of Moles

200 ml of C 2...

Question

200 ml of C₂H₄was burnt in sufficient air ( containing 20% oxygen) as per the equation C₂H₄+ 3O ₂ -------------> 2CO₂+ 2H₂O. Calculate the resultant mixture composition (at 100⁰C and constant pressure).

Open in App

Solution

C2H4 + 3O2 -> 2CO2 + 2H2O
Volume of ethene = 200 cm3
Volume of O2 required = 600 cm3
The amount of air = 600*100 /20 =3000cm3

Composition
Volume of CO2 & H2O = 400 cm3
Remaining volume of air = 3000 – 600 = 2400 cm3

Suggest Corrections

Similar questions

200 m l

C_{2} H_{4}

is burnt in just sufficient air (containing

20 %

oxygen). Calculate the resultant mixture composition at

100^{\circ} C

What volume of oxygen at STP is required to affect the combustion of 1 litres of ethylene $[C_{2} H_{4}]$ at $273^{\circ} C$ and 380 mm of Hg pressure?
$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$

How many grams of

C O_{2}

formed when 7.0 g of ethene

(C_{2} H_{4})

burns in oxygen?

C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O

Q. A mixture of

C H_{4}

and

C_{2} H_{4}

was completely burnt in excess of oxygen, yielding equal volumes of

C O_{2}

and steam. Calculate the percentages of the compounds in the original mixture:

Q. Air contains

21 %

of oxygen by volume. Calculate the volume of such air required for the complete combustion of

700 c m^{3}

of ethylene

(C_{2} H_{4})

15^{\circ} C

and

756 m m

pressure.

Join BYJU'S Learning Program

C2H4 + 3O2 -> 2CO2 + 2H2O Volume of ethene = 200 cm3 Volume of O2 required = 600 cm3 The amount of air = 600*100 /20 =3000cm3 Composition Volume of CO2 & H2O = 400 cm3 Remaining volume of air = 3000 – 600 = 2400 cm3

C2H4 + 3O2 -> 2CO2 + 2H2O
Volume of ethene = 200 cm3
Volume of O2 required = 600 cm3
The amount of air = 600*100 /20 =3000cm3

Composition
Volume of CO2 & H2O = 400 cm3
Remaining volume of air = 3000 – 600 = 2400 cm3