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)

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

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

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it

yes

ya te ayud espera