The reaction of methane and water is one way to prepare hydrogen:

CH4(g) + H2O(g) --> CO(g) + 3 H2(g)
[Molar masses: 16.04 18.02 28.01 2.02]

If you begin with 995 g of CH4 and 2510 g of water, what is the maximum possible yield of H2?

This is a limiting reagent problem.

Convert 995 g CH4 to moles.
Convert 2510 g H2O to moles.

Using the coefficients in the balanced equation, convert moles CH4 to moles H2.
Do the same and convert moles H2O to moles H2. Probably the number of moles will not agree which means one of them must be wrong; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Convert the smaller value of moles H2 to grams. g = moles x molar mass.

186×2=372g

no answer

To find the maximum possible yield of H2, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thus limiting the amount of product formed.

Step 1: Calculate the number of moles for CH4 and H2O.
To calculate the number of moles of a substance, divide its mass by its molar mass.

Number of moles of CH4 = 995 g / 16.04 g/mol = 61.993 mol
Number of moles of H2O = 2510 g / 18.02 g/mol = 139.311 mol

Step 2: Determine the stoichiometric ratio between CH4 and H2.
From the balanced equation, we can see that 1 mol of CH4 reacts to produce 3 mol of H2.

Step 3: Calculate the maximum possible yield of H2.
To find the maximum possible yield of H2, multiply the number of moles of CH4 by the stoichiometric ratio.

Maximum possible yield of H2 = 61.993 mol CH4 × (3 mol H2 / 1 mol CH4) = 185.979 mol H2

Step 4: Convert the maximum possible yield of H2 to grams.
To convert moles to grams, multiply the number of moles by the molar mass of H2.

Maximum possible yield of H2 = 185.979 mol H2 × 2.02 g/mol = 375.362 g H2

Therefore, the maximum possible yield of H2 is 375.362 grams.