Chemical Engineering - Chemical Reaction Engineering - Discussion

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%.

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%.

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%.

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 %.

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%.

The limiting reaction is carbon.

Any one tell me how can we find out answer?

@Arvind.

How can we get total moles of oxygen consumed?

Yes, you are right, Thanks @Madhuri.