# Question #0a4eb

##### 1 Answer

#### Explanation:

**!! EXTREMELY LONG ANSWER !!**

So, you know that you're dealing with an equilibrium reaction that involves gases, morespecifically methane,

I assume that the *equilibrium constant* for this reaction,

#K_p = 4.5 * 10^2#

since

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])#

By definition,

#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 **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%)#