the reaction of ethyl-tetrabromide with zinc dust proceed as shown in the figure
ID: 841847 • Letter: T
Question
the reaction of ethyl-tetrabromide with zinc dust proceed as shown in the figure below
the reaction is C_2H_2Br_4+2Zn-----> C_2H_2+2ZnBr_2 based on one pass through the reactor 80% of the C2H2Br4 is reacted and the remainder recycled on the basis of 1000kg of C2H2Br4 fed to the reactor per hour, calculate
a- How much C2H2 is produced per hour (in kg)
b-The rate of recycle in kg/hr
c- The feed rate necessary for Zn to be 20% in excess
d- The mole ratio of ZnBr2 to C2H2 in the final products
please do not copy the soultion from chegg answers , Thanks , for good answer 1500 points
Explanation / Answer
The balance equation which governs this reaction is as
C2H2Br4 + 2Zn ===> C2H2 + 2ZnBr2.
1Kmole(C2H2Br4) + 2Kmoles(Zn) ===>1Kmole(C2H2) + 2Kmoles(ZnBr2) (mole ratios in reaction)
346Kg/Kmole*1Kmole + 65.4kg/kmole*2Kmoles ==>26kg/kmole*1Kmole+ (225.4)Kg/kmole*2Kmoles (mass ratios)
346kg(C2H2Br4)+130.8kg(Zn) ===>26Kg(C2H2) + 450.8Kg(ZnBr2)
(since feed is 1000kg/hour C2H2Br4 divide all terms by 0.346*hour)
1000kg/hour(C2H2Br4) +378.03kg/hour(Zn) ===>75.14kg/hour(C2H2) + 1302.9kg/hour(ZnBr2) (for100% complete reaction)
(since 80% C2H2Br4 is reacted and remainder is recycled,multply all terms of equation by 0.8)
800kg/hour(C2H2Br4) +302.42kg/hour(Zn)===>60.11kg/hour(C2H2) + 1042.32kg/hour(ZnBr2)
A) How much C2H2 is produced per hour (in kg).
60.11kg/hour(C2H2) assuming other reactant is in excess and *)% goes to completion
B) The rate of recycle in kg/hr.
Rate of Recycle(C2H2Br4)=Feed Rate -Recycle rate=(1000-800)=200kg/hour
C) The feed rate necessary for Zn to be 20% excess
Excess feed rate of Zn=(Feed rate for 100% reaction-feed rate for 80%)
=(378.03-302.42)=75.61kg/hour Zn in excess
D) The mole ratio of ZnBr2 to C2H2 in the final products.
Mole ratio in products is the same as in stiochiometric equation 1mole(C2H2):2mole(ZnBr2) or 1:2
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