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What volume of oxygen at STP is needed to fully react with 8.53 mol of C2H4?
C2H4 reacts with O2, according to the following equation:

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)

Respuesta :

Answer: 574 L

Explanation: Solution attached:

First find the number of moles of O2 using the mole ratio between C2H4

Next to find Volume use PV=nRT or ideal gas law.

Derive for V:

V= nRT/P

Substitute the values

At STP: Pressure is 1 atm, Temperature is 273K

25.59moles O2( 0.0821L.atm/mol.K)

( 273K)/ 1 atm

= 574 L

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Eduard22sly

Answer: 573.22L

Explanation:

C2H4 + 3O2 → 2CO2 + 2H2O

From the equation,

1mole of C2H4 reacted with 3 moles of O2.

Therefore, 8.53 mol of C2H4 will react with = 8.53x3 = 25.59moles of O2.

But 1mole of O2 contains 22.4L at stp.

Therefore, 25.59moles will contain = 25.59 x 22.4 = 573.22L

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