The reaction of methane and water is one way to prepare hydrogen: CH_4(g) + H_2O (g) -> CO (g) + 3H_2(g) [Molar masses 16.04 18.02 28.01 2.02]. If you begin with 995 g of CH_4 and 2510 g of water, what is the maximum possible yield of H_2?

Aug 1, 2017

Methane is the limiting reactant, therefore the maximum possible yield of hydrogen gas is 376 g.

Explanation:

Balanced Equation

$\text{CH"_4("g") + "H"_2"O(g)}$$\rightarrow$$\text{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.

$\textcolor{b l u e}{\text{Maximum Yield of}}$ color(blue)("H"_2" color(blue)("Produced by" color(blue)("995 g CH"_4

Determine the moles ${\text{CH}}_{4}$ in ${\text{995 g CH}}_{4}$ 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 ${\text{CH}}_{4}$ by the mole ratio between ${\text{H}}_{2}$ and ${\text{CH}}_{4}$, ${\text{3 mol H}}_{2} :$${\text{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"

$\textcolor{t e a l}{\text{Maximum Yield of}}$ $\textcolor{t e a l}{{\text{H}}_{2}}$ color(teal)("produced by" color(teal)(2510"g H"_2"O"

Determine the moles $\text{H"_2"O}$ 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 $\text{H"_2"O}$ by the mole ratio between ${\text{H}}_{2}$ and $\text{H"_2"O}$, ${\text{3 mole H}}_{2} :$$1 \text{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 ${\text{CH}}_{4}$ since it produced the least number of moles of ${\text{H}}_{2}$. In order to determine the mass $\text{H"_2}$ that is possible, multiply the moles ${\text{H}}_{2}$ by the molar mass of ${\text{H}}_{2}$.

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)