Flows into a catalytic reactor at 26.2 atm and 250.°c with a flow rate of 1100. l/min. hydrogen at 26.2 atm and 250.°c flows into the reactor at a flow rate of 1400. l/min. if 13.9 kg is collected per minute, what is the percent yield of the reaction?

69%

Explanation:

Say the hydrocarbon is C2H4, the equation of reaction would be;

C2H4 + H2 > C2H6

The hydrocarbon, C2H4 is the limiting reagent.

From the question, the hydrocarbon C2H4 flows into a catalytic reactor at 26.2 atm and 250°C with a flow rate of 1100 L/min

Using, PV=nRT(1).

n= PV/RT to find the number of moles.

n= 26.2 atm × 1100 L/ 0.0821 L. atm/mol per kelvin × 523 K.

=28,820 atm.L/ 42.9383 L.atm/mol.

= 671.2 mole.

One mole of C2H4 produced one mole of C2H6.

Mass of C2H6 = 30 × 671.2

= 20,136 g = 20.136 kg.

Percent yield = actual yield/ theoretical yield × 100% (2)

Actual yield= 13.9 kg, theoretical yield = 20.136 kg

Substitute the parameters into equation (2). We have;

Percent yield = 13.9 kg/ 20.136× 100

Percent yield= 0.69 × 100

Percent yield= 69%

Note: P= pressure, V= volume, T= temperature and n = number of moles

i think it is true since you can test the answer and can use observations to complete it.

i would say b

explanation:

but i think i would need more information than that