posted by Rachel :) on .
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.