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 $H_2$ ?

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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 $H_2$ ?

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Answer 1 · 2017-08-01T09:25:43

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

Explanation:

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)

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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 $H_2$ ?

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The reaction of methane and water is one way to prepare hydrogen: CH_4(g) + H_2O (g) -> CO (g) + 3H_2(g)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_4CH_4 and 2510 g of water, what is the maximum possible yield of H_2H_2?

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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 $H_2$ ?