Balanced Equation
#"CH"_4("g") + "H"_2"O(g)"##rarr##"CO(g)" + "3H"_2("g")"#
This is a limiting reactant stoichiometry problem. We will determine whether methane or water is the limiting reactant. The limiting reactant will give us the maximum possible yield of hydrogen gas.
#color(blue)("Maximum Yield of")# #color(blue)("H"_2"# #color(blue)("Produced by"# #color(blue)("995 g CH"_4#
Determine the moles #"CH"_4# in #"995 g CH"_4# by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.
#995color(red)cancel(color(black)("g CH"_4))xx(1"mol CH"_4)/(16.04color(red)cancel(color(black)("g CH"_4)))="62.0 mol CH"_4#
Multiply the moles #"CH"_4# by the mole ratio between #"H"_2# and #"CH"_4#, #"3 mol H"_2:##"1 mol CH"_4#.
#62.0color(red)cancel(color(black)("mol CH"_4))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol CH"_4)))="186 mol H"_2"#
#color(teal)"Maximum Yield of"# #color(teal)("H"_2)# #color(teal)("produced by"# #color(teal)(2510"g H"_2"O"#
Determine the moles #"H"_2"O"# by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.
#2510color(red)cancel(color(black)("g H"_2"O"))xx(1"mol H"_2"O")/(18.02color(red)cancel(color(black)("g H"_2"O")))="139 mol H"_2"O"#
Multiply the moles #"H"_2"O"# by the mole ratio between #"H"_2# and #"H"_2"O"#, #"3 mole H"_2:##1"mol H"_2"O"#.
#139color(red)cancel(color(black)("mol H"_2"O"))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol H"_2"O")))="417 mol H"_2"#
The limiting reactant is #"CH"_4# since it produced the least number of moles of #"H"_2#. In order to determine the mass #"H"_2"# that is possible, multiply the moles #"H"_2# by the molar mass of #"H"_2#.
#color(magenta)("Maximum Possible Yield of H"_2"#
#186color(red)cancel(color(black)("mol H"_2))xx(2.02"g H"_2)/(1color(red)cancel(color(black)("mol H"_2)))="376 g H"_2"# (rounded to three sig figs)
