The reaction of methane and water is a way to prepare hydrogen according to the

ID: 838131 • Letter: T

Question

The reaction of methane and water is a way to prepare hydrogen according to the equation:

CH4(g)+H2O(l)-->CO2(g)+H2(g) If 995g of CH4 reacts with 2510g of water:

a) identify the limiting reagent in the reaction

b) how many moles of reactants (CH4(g)&H20(l)) and products (CO2(g)&H2(g)) are there when the reaction is finished?

c) if only 228g of H2 are produced, what is the % yield of the gas?

Also i dont understand how to find the inital change and final for each part

CH4(g) + H2O --> CO2(g) + H2 (g)

Initial:

Change:

Final:

Balanced reaction is given by CH4 +2H2O - CO2 + 4H2

moles of CH4 = 995/16 = 62.1875 moles of H2O = 2510/18 = 139.45

1 mol Ch4 reacts with 2 mol H2O 62.1875 reacts with 124.375 mol of water

Since water is in excess CH4 IS LIMITING REAGANT

When reaction is finished Ch4 is consumed completely

Moles of water left = 139.45 - 124.375 = 15.075

Moles of CO2 = 62.1875

Moles of H2 = 248.75

grams of H2 produced theoretically = 497.5 gm actual yield = 228 gm

% yield = 228/497.5 x 100 = 45.829%

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