n the water gas shift reaction (below, unbalanced), equal masses of methane and water were placed in
the reactor and the total mass of the reactants was 274 kg.
_______ CH4 (g) + _______ H2O (g) → _______H2 (g) + _______ CO2 (g)
Given: MW of CH4 is 16.04 g/mol, MW of H2O is 18.01 g/mol, MW of H2 is 2.02 g/mol
So just a quick q, would I assume that eah reactants starting amount is 137kg and work from there to find moles
10 answers
after balancing ofcourse
Also can someone check if my answers are right
a) Calculate the moles of CH4(g) and H2O(g) in the reactor.
c) What is the limiting reactant?
d)
If the reaction is found to have a 73.2% yield, what actual mass of hydrogen was
obtained?
So for c)I got H20 d) I got 200386?
a) Calculate the moles of CH4(g) and H2O(g) in the reactor.
c) What is the limiting reactant?
d)
If the reaction is found to have a 73.2% yield, what actual mass of hydrogen was
obtained?
So for c)I got H20 d) I got 200386?
Yes, 137 g H2O and 137 g CH4 initially.
Yes, H2O is the limiting reagent.
However, I don't get your answer for kg H2 produced at 73.2% yield
However, I don't get your answer for kg H2 produced at 73.2% yield
What did you get?
Wait so I got 11136 g or 11.136 (not really caring about sig figs atm)
My answer is about 22.5 kg H2 produced at 73.2% yield..
mol H2O= 7.61
mols CH4 = 8.54
mols H2 produced = 7.61 x 2 = about 15
g H2 = mols H2 x 2.02 x 0.732 = 22.5
Note: I did NOT calculate mols so those numbers above for mols et al are not real; i.e., I did not convert kg to g and back to kg. I just used the molar mass as kg; i.e., 16.04, 18.01, 2.02 etc as kg/mol instead of g/mol. That makes the answer come out in kg. If you converted first, and I think you probably did that, then my mols CH4, mols H2O and so forth will be 1000 times what I wrote above.
mol H2O= 7.61
mols CH4 = 8.54
mols H2 produced = 7.61 x 2 = about 15
g H2 = mols H2 x 2.02 x 0.732 = 22.5
Note: I did NOT calculate mols so those numbers above for mols et al are not real; i.e., I did not convert kg to g and back to kg. I just used the molar mass as kg; i.e., 16.04, 18.01, 2.02 etc as kg/mol instead of g/mol. That makes the answer come out in kg. If you converted first, and I think you probably did that, then my mols CH4, mols H2O and so forth will be 1000 times what I wrote above.
Did yu balance the equation first.
That could account for the factor of 2.
CH4 + 2H2O ==> CO2 + 4H2
That could account for the factor of 2.
CH4 + 2H2O ==> CO2 + 4H2
hmm.Yeah i balanced it
CH4 + 2H2O ==> CO2 + 4H2
274 kg
137 kg CH4
137 kg H2O
mols CH4 = 137,000/16.04 = 8,541
mols H2O = 137,000/18.01 = 7,607
We agree H2O is the limiting reagent; therefore, mols H2 produced is
7607 mols H2O x (4 mols H2/2 mols H2O) = 15,214.
g H2 @ 100% yield = mols H2 x molar mass H2 = 15,214 x 2.02 = 30,732 g
g H2 @ 73.2% yield = 30,732 x 0.732 = 22,496 g = 22.5 kg.
If you plug in all the numbers and don't take the numbers out of the calculator (taking them out leads to rounding errors) it looks like this
(137,000/18.01) x 2 x 2.02 x 0.732 = 22,496 g = 22.496 kg but that's too many s.f. I'd still report it as 22.5 kg. Let me know if you see an error. .
274 kg
137 kg CH4
137 kg H2O
mols CH4 = 137,000/16.04 = 8,541
mols H2O = 137,000/18.01 = 7,607
We agree H2O is the limiting reagent; therefore, mols H2 produced is
7607 mols H2O x (4 mols H2/2 mols H2O) = 15,214.
g H2 @ 100% yield = mols H2 x molar mass H2 = 15,214 x 2.02 = 30,732 g
g H2 @ 73.2% yield = 30,732 x 0.732 = 22,496 g = 22.5 kg.
If you plug in all the numbers and don't take the numbers out of the calculator (taking them out leads to rounding errors) it looks like this
(137,000/18.01) x 2 x 2.02 x 0.732 = 22,496 g = 22.496 kg but that's too many s.f. I'd still report it as 22.5 kg. Let me know if you see an error. .