The feed to a propylene ammoxidation process contains 14.55 % of C3H, by mass, 7.0 % ammonia by mass, and 78.45 % air by mass. A fractional conversion of 30.0% of the limiting reactant is achieved, taking 1000 kg feed as a basis, determine the ratio of the unreacted oxygen to the feed by mass, the percentage by which each of the other reactions is in excess, and the mass amounts of all product gas constituents for a 30% conversion of the limiting reactant Feed = 100 mole P= ? Reactor x = 0.1455 NNH, = 0.07 nor xar = 0.7845 =? =? =? C3H, + NH3 +3/20, C3H3N +3H20
The feed to a propylene ammoxidation process contains 14.55 % of C3H, by mass, 7.0 % ammonia by mass, and 78.45 % air by mass. A fractional conversion of 30.0% of the limiting reactant is achieved, taking 1000 kg feed as a basis, determine the ratio of the unreacted oxygen to the feed by mass, the percentage by which each of the other reactions is in excess, and the mass amounts of all product gas constituents for a 30% conversion of the limiting reactant Feed = 100 mole P= ? Reactor x = 0.1455 NNH, = 0.07 nor xar = 0.7845 =? =? =? C3H, + NH3 +3/20, C3H3N +3H20
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![e A:Y
* Asiacell liı.
1st month 2nd...
Q1
The feed to a propylene ammoxidation process contains 14.55 % of C3H, by
mass, 7.0 % ammonia by mass, and 78.45 % air by mass. A fractional conversion
of 30.0% of the limiting reactant is achieved, taking 1000 kg feed as a basis,
determine the ratio of the unreacted oxygen to the feed by mass, the percentage
by which each of the other reactions is in excess, and the mass amounts of all
product gas constituents for a 30% conversion of the limiting reactant
Feed = 100 mole
P= ?
Reactor
x = 0.1455
NNH, =?
x = 0.07
no:
=?
xhir = 0.7845
=?
=?
C3H, + NH3 + 3/202 C3H,N + 3H20
Q2
Sulphur trioxide gas is obtained by the combustion of pyrites (FeS2) according
to the following reaction:
4FES2 + 1502 – 2Fe,03 + 85O3
The reaction is accompanied by the following side reaction:
4FES2 + 1102 - 5FE203 + 8SO2
Assume that 80% (weight) of the pyrites charged reacts to give sulphur trioxide
and 20% reacts giving sulphur dioxide.
a) How many kilograms pf pyrites charged will give 100 kg of SO3?
b) How many kilograms of oxygen will be consumed in the reaction?
Mwt:
C3H6 = 42, NH3 = 17,02 = 32, C3H3N = 53, H20 = 18, N2 = 28, FeS2 = 120
Fe,0, = 136, SO, 80, SO, = 64](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa90faa31-a6c2-42f8-ba51-49b17a1e9d6e%2F39eb0ddd-7585-4916-8eee-150645a78b6b%2Fizkdyl9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:e A:Y
* Asiacell liı.
1st month 2nd...
Q1
The feed to a propylene ammoxidation process contains 14.55 % of C3H, by
mass, 7.0 % ammonia by mass, and 78.45 % air by mass. A fractional conversion
of 30.0% of the limiting reactant is achieved, taking 1000 kg feed as a basis,
determine the ratio of the unreacted oxygen to the feed by mass, the percentage
by which each of the other reactions is in excess, and the mass amounts of all
product gas constituents for a 30% conversion of the limiting reactant
Feed = 100 mole
P= ?
Reactor
x = 0.1455
NNH, =?
x = 0.07
no:
=?
xhir = 0.7845
=?
=?
C3H, + NH3 + 3/202 C3H,N + 3H20
Q2
Sulphur trioxide gas is obtained by the combustion of pyrites (FeS2) according
to the following reaction:
4FES2 + 1502 – 2Fe,03 + 85O3
The reaction is accompanied by the following side reaction:
4FES2 + 1102 - 5FE203 + 8SO2
Assume that 80% (weight) of the pyrites charged reacts to give sulphur trioxide
and 20% reacts giving sulphur dioxide.
a) How many kilograms pf pyrites charged will give 100 kg of SO3?
b) How many kilograms of oxygen will be consumed in the reaction?
Mwt:
C3H6 = 42, NH3 = 17,02 = 32, C3H3N = 53, H20 = 18, N2 = 28, FeS2 = 120
Fe,0, = 136, SO, 80, SO, = 64
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