Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 1 (Q.No. 15)
15.
6 gm of carbon is burnt with an amount of air containing 18 gm oxygen. The product contains 16.5 gms CO2 and 2.8 gms CO besides other constituents. What is the degree of conversion on the basis of disappearance of limiting reactant ?
Discussion:
9 comments Page 1 of 1.
Chalachew nigussie said:
5 years ago
The balanced chemical equation is 4C+3O2 gives 2CO2+2CO.
Then mole of O2 = 18g/32(g/mol)= 0.5625 mole,
mole of C = 6g/12(g/mol) = 0.5 mole, so from the balanced equation for 3 mole of O2 4 mole of C is required and for 0.5625 mole of O2 = (0.5625*4)/3 = 0.75 mole of C is required.
But we have 0.5 mole of C. So the limiting reactant is carbon. mole of CO2 produced = 16.5g/44(g/mol) =0.375 mole from which mole of carbon consumed is 0.375*1=0.375 mole.
And mole of CO produced =2.8/28 = 0.1 mol from which mole of C consumed = 0.1*1 = 0.1 mole. the total C consumed is 0.375+0.1 = 0.475 mole.
So conversion = (0.475/0.5)*100% = 95%.
Then mole of O2 = 18g/32(g/mol)= 0.5625 mole,
mole of C = 6g/12(g/mol) = 0.5 mole, so from the balanced equation for 3 mole of O2 4 mole of C is required and for 0.5625 mole of O2 = (0.5625*4)/3 = 0.75 mole of C is required.
But we have 0.5 mole of C. So the limiting reactant is carbon. mole of CO2 produced = 16.5g/44(g/mol) =0.375 mole from which mole of carbon consumed is 0.375*1=0.375 mole.
And mole of CO produced =2.8/28 = 0.1 mol from which mole of C consumed = 0.1*1 = 0.1 mole. the total C consumed is 0.375+0.1 = 0.475 mole.
So conversion = (0.475/0.5)*100% = 95%.
(2)
Madhuri said:
1 decade ago
C (12g) + O2(32g) = CO2(44g).
C(12g) + O(16g) = CO(28g).
For 6 g carbon we need 16 gm oxygen for CO2.
And 8 gm oxygen for CO.
Hence carbon is limiting,
For 16.5 gm CO2 , carbon used = 16.5*12/44.
= 4.5 gm.
For 2.8 gm CO , carbon used = 2.8*12/28.
= 1.2.
Total carbon reacted is 4.5+1.2 = 5.7.
Conversion = ( 5.7/6)*100 = 95%.
C(12g) + O(16g) = CO(28g).
For 6 g carbon we need 16 gm oxygen for CO2.
And 8 gm oxygen for CO.
Hence carbon is limiting,
For 16.5 gm CO2 , carbon used = 16.5*12/44.
= 4.5 gm.
For 2.8 gm CO , carbon used = 2.8*12/28.
= 1.2.
Total carbon reacted is 4.5+1.2 = 5.7.
Conversion = ( 5.7/6)*100 = 95%.
(4)
Arvind kumar tiwari said:
7 years ago
C + O2 = CO2 --> 1st.
C + O= CO --> 2nd.
Given.
Carbon -6gm - .5 mol.
O2 -18gm -1.125mol.
From reaction 1.
Moles of CO2 produce=moles of C consumd = .375 mol.
From reaction 2.
Moles of CO produce=moles of C consumd = .1moles.
Total moles of Corbon consumd=.457.
Conversion = (.457/.5)*100=95%.
C + O= CO --> 2nd.
Given.
Carbon -6gm - .5 mol.
O2 -18gm -1.125mol.
From reaction 1.
Moles of CO2 produce=moles of C consumd = .375 mol.
From reaction 2.
Moles of CO produce=moles of C consumd = .1moles.
Total moles of Corbon consumd=.457.
Conversion = (.457/.5)*100=95%.
(1)
Sumit Kumar Rao said:
3 years ago
1C + O2 = 1CO2.
1C + 1/2O2 = 1CO.
Co2 - 16.5grm( n=16.5/44= .375).
CO - 2.8grm {n = 2.8/28 = .1}.
1C - 1CO2.
XC- .375co2.
X=.375.
1C-1Co.
XC- .1co.
Y= .1.
X + Y = .475.
C used = .475.
6grm of C in , n = 6/12 =.5.
%C = .475/.5 * 100 = 95%.
1C + 1/2O2 = 1CO.
Co2 - 16.5grm( n=16.5/44= .375).
CO - 2.8grm {n = 2.8/28 = .1}.
1C - 1CO2.
XC- .375co2.
X=.375.
1C-1Co.
XC- .1co.
Y= .1.
X + Y = .475.
C used = .475.
6grm of C in , n = 6/12 =.5.
%C = .475/.5 * 100 = 95%.
(6)
Sateesh meena said:
1 decade ago
Limiting Reactant is O2.
Mole of O2 consumed= (mole of CO2 formed + mole of CO of formed/2)= 16.5/44 + 2.8/(28*2) = 0.425.
Mole of O2 supplied = 18/32.
Conversion = 0.425/(18/32)*100 = 75.5 %.
Mole of O2 consumed= (mole of CO2 formed + mole of CO of formed/2)= 16.5/44 + 2.8/(28*2) = 0.425.
Mole of O2 supplied = 18/32.
Conversion = 0.425/(18/32)*100 = 75.5 %.
(1)
Dave said:
6 years ago
@Arvind.
How can we get total moles of oxygen consumed?
How can we get total moles of oxygen consumed?
Navdeep said:
1 decade ago
Any one tell me how can we find out answer?
Suyash said:
4 years ago
Yes, you are right, Thanks @Madhuri.
Kasahun said:
4 years ago
The limiting reaction is carbon.
(1)
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