Question #0a4eb

!! EXTREMELY LONG ANSWER !!

So, you know that you're dealing with an equilibrium reaction that involves gases, morespecifically methane, #"CH"_4#, carbon dioxide, #"CO"_2#, carbon monoxide, #"CO"#, and hydrogen, #"H"_2#.

I assume that the equilibrium constant for this reaction, #K_p#, is actually

#K_p = 4.5 * 10^2#

since #4.5^2# doesn't make much sense to me.

The idea here is that you need to find the moles of methane and carbon dioxide that are present in the container. The use the ideal gas law equation - notice that they tell you to assume ideal gas behavior - to find the partial pressures of methane and carbon dioxide.

Next, use an ICE table to find the partial pressure of hydrogen.

So, the number of moles of methane and carbon dixode are

#22.3color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole CH"""_4)/(16.04color(red)(cancel(color(black)("g")))) = "1390.3 moles CH"""_4#

and

#55.4color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole CO"""_2)/(44.01color(red)(cancel(color(black)("g")))) = "1258.8 moles CO"""_2#

The partial pressures of these ttwo gases will be

#PV = nRT implies P = (nRT)/V#

#P_(CH_2) = (1390.3color(red)(cancel(color(black)("moles"))) * 0.082("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 825color(red)(cancel(color(black)("K"))))/(85color(red)(cancel(color(black)("L")))) = "1106.5 atm"#

and

#P_(CO_2) = (1258.8color(red)(cancel(color(black)("moles"))) * 0.082("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 825color(red)(cancel(color(black)("K"))))/(85color(red)(cancel(color(black)("L")))) = "1001.9 atm"#

Now use an ICE table to find the partial pressure of hydrogen

#"CH"_text(4(g]) " "+" " "CO"_text(2(g]) " "rightleftharpoons" " color(red)(2)"CO"_text((g]) " "+" " color(red)(2)"H"_text(2(g])#

#color(purple)("I")" " "1106.5" " " " " " "1001.9" " " " " " " " 0" " " " " " " " " "0#
#color(purple)("C")" " (-x)" " " " " " " " "(-x)" " " " " " " (+color(red)(2)x)" " " " " " "(+color(red)(2)x)#
#color(purple)("E")" (1106.5-x)" " " " (1001.9-x)" " " " "(color(red)(2)x)" " " " " " " "(color(red)(2)x)#

By definition, #K_p# will be

#K_p = ( ("CO")^color(red)(2) * ("H"_2)^color(red)(2))/(("CH"_4) * ("CO"_2)) = ( (2x)^2 * (2x)^2)/((1106.5 - x)(1001.9 - x))#

#K_p = (16x^4)/((1106.5 - x)(1001.9 - x)) = 4.5 * 10""^2#

Now, provided that you don't want a major headache, you will use

#(1106.5-x) ~~ 1106.5#
#(1001.9 - x) ~~ 1001.9#

This will get you

#x = root(4)((4.5 * 10^2 * 1106.5 * 1001.9)/16) = 74.725#

The equilibrium partial pressure of hydrogen gas will be

#P_(H_2) = 2 * x = 2 * 74.725 = "149.45 atm"#

Now use the ideal gas law equation again to find the number of moles of hydrogen gas you have in the container at equilibrium

#n = (PV)/(RT)#

#n_(H_2) = (149.45color(red)(cancel(color(black)("atm"))) * 85color(red)(cancel(color(black)("L"))))/(0.082("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 825color(red)(cancel(color(black)("K")))) = "187.78 moles H"""_2#

Use hydrogen gas' molar mass to determine how many grams would contain this many moles

#187.78color(red)(cancel(color(black)("moles H"""_2))) * ("2.016 g H"""_2)/(1color(red)(cancel(color(black)("mole H"""_2)))) = "378.6 g H"""_2#

I will round this off to two sig figs, the number of sig figs you gave for the volume of the container

#m_(H_2) = color(green)("380 g")#

Now, to get the percent yield of the reaction under these conditions, use the mole ratios that exists between the species involved in the reaction to get the theoretical yield.

Notice that you have a #1:1# mole ratio between methane and carbon dioxide. This means that you need equal numbers of moles of each in order for the reaction to take place.

Since you have more moles of methane than you have of carbon dioxide, the latter will act as a limiting reagent.

The number of moles of hydrogen produced by the reaction will theoretically be

#1001.9color(red)(cancel(color(black)("moles CO"""_2))) * (color(red)(2)" moles H"""_2)/(1color(red)(cancel(color(black)("mole CO"""_2)))) = "2003.8 moles H"""_2#

The mass of hydrogen will be

#2003.8color(red)(cancel(color(black)("moles H"""_2))) * ("2.016 g H"""_2)/(1color(red)(cancel(color(black)("mole H"""_2)))) = "4039.7 g H"""_2#

The percent yield of the reaction will thus be

#"% yield" = "actual yield"/"theoretical yield" xx 100#

#"% yield" = (380color(red)(cancel(color(black)("g"))))/(4039.7color(red)(cancel(color(black)("g")))) xx 100 = color(green)(9.4%)#