Ethene is converted to ethane by the reaction c2h4 1g2 1 h2 1g2 n c2h6 1g2 c2h4 flows into a catalytic reactor at 25.0 atm and 300.8c with a flow rate of 1000. l/min. hydrogen at 25.0 atm and 300.8c flows into the reactor at a flow rate of 1500. l/min. if 15.0 kg c2h6 is collected per minute, what is the percent yield of the reaction

Percentage Yield = 93.8%

Explanation:

Data:

R = 0.08206 L.atm/mol/K,

Molar mass of C2H6 =  30.08 g/mol

T = 300° c +273K = 573.15 K

C2H4 Conditions,

P = 25 atm

V = 1000 L/min

H2 Condition,

P = 25 atm

V = 1500 L/min

Solution:

First we will calculate the moles of ethane produced in one minute,

nC2H6 = (15/30.08) ( 1000/1)

nC2H6 = 498.67 mol/min

Now we will calculate the moles of ethane flowing in one minute using the ideal gas equation which is, Pv = nRT

n = pV/R

nC2H4 = ((25* 1000)/(0.08206*573.15))

nC2H4 = 531.54 mol/min

Similarly  Calculating the moles of hydrogenflowing in one minute,

nH2 = ((25*1500)/(0.08206)(573.15))

nH2 = 797.32 mol/mim

Both hydrogen and ethene are flowing at 25 atm and 573.15 K but the flow rate of hydrogen is more than that of ethane which results in less amount of available ethene for the reaction. We know that 1 mole of ethene reacts with 1 mole of hydrogen to give 1 mole of ethane

hence ethene is acting as the limiting reagent here

Now calculating the %Yield = (498.67/531.54) * 100

%Yield = 93.82%

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3