Mass (kg) H, usage(kg/s) = Time (s) Then we have: 2.02(g) (kg) Mass (kg) = Moles (mol)- Moles (mol) 1000(g) And we also have: 60 (s) Time (s) = Minutes(min). (min) 60Minutes Substituting into the left hand side of the formula for hydrogen usage, we have: Mass (kg) Moles Moles H, usage (kg/s) = Time (s) 60Minutes Minutes Inserting into the full formula for hydrogen usage, we have: P,(W) V,(V) H, usage(mol/min)
Mass (kg) H, usage(kg/s) = Time (s) Then we have: 2.02(g) (kg) Mass (kg) = Moles (mol)- Moles (mol) 1000(g) And we also have: 60 (s) Time (s) = Minutes(min). (min) 60Minutes Substituting into the left hand side of the formula for hydrogen usage, we have: Mass (kg) Moles Moles H, usage (kg/s) = Time (s) 60Minutes Minutes Inserting into the full formula for hydrogen usage, we have: P,(W) V,(V) H, usage(mol/min)
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter17: Electrochemistry And Its Applications
Section: Chapter Questions
Problem 83QRT
Related questions
Question
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Answer those blank. Show complete solutions.
![(c) Consider the equation for hydrogen consumption mµ2 in a fuel cell:
P,(W)
H, usage(kg/s) =1.05×10*
V (V)
Convert this equation to one where hydrogen usage is reported in mol/hr.
Jason M. Keith
Student View
Supplemental Material for Elementary Principles of Chemical Processes
Strategy
We can use unit conversions to determine a new equation for the hydrogen consumption
rate.
Solution
Start with the equation and the given units, writing the usage rate as a ratio of mass per
unit time.
Mass (kg)
H, usage(kg/s) =
Time (s)
Then we have:
2.02(g) (kg)
Mass (kg) = Moles (mol)-
Moles
(mol) 1000(g)
And we also have:
60 (s)
Time (s) = Minutes(min).
= 60Minutes
(min)
Substituting into the left hand side of the formula for hydrogen usage, we have:
Mass (kg)
Moles
Moles
H¸ usage(kg/s) =
Time (s)
60Minutes
Minutes
Inserting into the full formula for hydrogen usage, we have:
P,(W)
H, usage(mol/min) =
V,(V)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F716ad723-1111-43dd-b3c4-d535f081a762%2F61724f1e-0106-4c44-bab6-ac3d8eb54d22%2Fxfkw17_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(c) Consider the equation for hydrogen consumption mµ2 in a fuel cell:
P,(W)
H, usage(kg/s) =1.05×10*
V (V)
Convert this equation to one where hydrogen usage is reported in mol/hr.
Jason M. Keith
Student View
Supplemental Material for Elementary Principles of Chemical Processes
Strategy
We can use unit conversions to determine a new equation for the hydrogen consumption
rate.
Solution
Start with the equation and the given units, writing the usage rate as a ratio of mass per
unit time.
Mass (kg)
H, usage(kg/s) =
Time (s)
Then we have:
2.02(g) (kg)
Mass (kg) = Moles (mol)-
Moles
(mol) 1000(g)
And we also have:
60 (s)
Time (s) = Minutes(min).
= 60Minutes
(min)
Substituting into the left hand side of the formula for hydrogen usage, we have:
Mass (kg)
Moles
Moles
H¸ usage(kg/s) =
Time (s)
60Minutes
Minutes
Inserting into the full formula for hydrogen usage, we have:
P,(W)
H, usage(mol/min) =
V,(V)
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