I've found the number of moles of each gas (H2 and H2O) using PV = nRT. Now. I need to find the theoretical yield of H at STP. Can you refresh my memory on how to do that? Is it different with gases?

#CH_4#(g) + #H_2O#(g) #rarr# CO(g) + 3#H_2#(g)

In a particular reaction, 25.5L of methane gas (pressure 732 torr, temp 25C) mixes with 22.8 L water vapor (pressure 702 torr, temp 25C)

1 Answer
Nov 5, 2015

Answer:

#"2.58 moles H"_2#

Explanation:

SIDE NOTE In order to make sure that other students can benefit from this as well, I'll solve the problem completely. If you already found the number of moles of hydrogen and of water, you can just skip to the middle of the answer.

Your starting point here will be the balanced chemical equation for this reaction

#"CH"_text(4(g]) + "H"_2"O"_text((g]) -> "CO"_text((g]) + color(red)(3)"H"_text(2(g])#

Now, the first thing to do here is determine whether or not you're dealing with a limiting reagent. Notice that you have a #1:1# mole ratio between methane and water.

This means that the reaction will consume equal numbers of moles of each reactant. To determine how many moles of each reactant you have, use the ideal gas law equation

#color(blue)(PV = nRT)" "#, where

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant, usually given as #0.082("atm" * "L")/("mol" * "K")#
#T# - the temperature of the gas, expressed in Kelvin

Plug in your values and solve for #n#, the number of moles of gas, to get

#"For CH"_4: " "n = (732/760color(red)(cancel(color(black)("atm"))) * 25.5color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))#

#n = "1.0046 moles CH"_4#

#"For H"_2"O": " " n = (702/760color(red)(cancel(color(black)("atm"))) * 22.8color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))#

#n = "0.8614 moles H"_2"O"#

Notice that you don't have enough moles of water to ensure that all the methane will react. This means that water will be the limiting reagent, or, looking at this from the other perspective, methane will be in excess.

Now, to get the theoretical yield of hydrogen at STP, Standard Temperature and Pressure, you need to use the #1:color(red)(3)# mole ratio that exists between either methane or hydrogen, or water and hydrogen.

Since water determines how many moles of methane react, your reaction will produce

#0.8614color(red)(cancel(color(black)("moles H"_2"O"))) * (color(red)(3)" moles H"_2)/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "2.584 moles H"_2#

This will be your reaction's theoretical yield, i.e. the amount of hydrogen produced if the reaction has a #100%# yield.

Normally, you'd have to round this off to two sig figs, the number of sig figs you have for the temperature of the reactants, but I'll leave the answer rounded to three sig figs

#n_(H_2) = color(green)("2.58 moles H"_2)#

From this point' you'll probably have to use the reaction's actual yield to determine its percent yield.

If they give you a volume of hydrogen, use the STP conditions and the molar volume of a gas at STP to determine how many moles were actually produced.

For example, let's sy that the reaction produced #"47.9 L"# of hydrogen gas at STP. You know that STP conditions imply a pressure of #"100 kPa"# and a temperature of #0^@"C"#.

Under these conditions, one mole of ny ideal gas occupies exactly #"22.7 L"#. This means that the reaction produced

#47.9color(red)(cancel(color(black)("L"))) * "1 mole H"_2/(22.7color(red)(cancel(color(black)("L")))) = "2.11 moles H"_2#

The percent yield of the reaction would thus be

#color(blue)("%yield" = "actual yield"/"theoretical yield" xx 100)#

#"% yield" = (2.11color(red)(cancel(color(black)("moles"))))/(2.58color(red)(cancel(color(black)("moles")))) xx 100 = 81.8%#