Question

1 - Mass balance with reaction

a) A certain amount of Pennsylvania coal, in a final analysis (by mass), has 84.36% C, 89% H2, 4.40% O2, 0.63% N2, 0.89% S and 7.83% ash (non-combustible). This coal is burned with a theoretical amount of air. Disregarding the ash content, determine the molar fractions of the products.

84,36% C 1,89% H2 4,40% O2 Carvão 0,63% N2 0,89% S 7,83% Cinzas Gases do produto Câmara de combustão Cinzas Ar teórico

b) The antimony is obtained by heating the tibite (Sb₂S₃) sprayed with iron shavings, removing the molten antimony from the bottom of the reaction tank:

Sb₂S₃ + 3Fe → 2Sb + 3FeS

Suppose 0.600 kg of sibite and 0.250 kg of iron shavings are heated together, producing 0.200 kg of metallic Sb. Determine a. The limiting reagent

1. The percentage of excess reagent
2. The extent of the reaction
3. The conversion percentage based on Sb₂S₃
4. Sb yield in kilograms produced per kilogram of Sb₂S₃ fed into reactor

c) In anaerobic fermentation of grains, the yeast Saccharomyces cerevisiae digests the glucose of the vegetable to form the products ethanol and propaneic acid, according to the following global reactions:

Reaction 1: C₆H₁₂O₆ → 2C₂H₅OH +2CO₂

Reaction 2: C₆H₁₂O₆ → 2C₂H₃CO₂H +2H₂O

In an open reactor, 3500 kg of a 12% glucose aqueous solution flows into the reactor. During fermentation, 120 kg of carbon dioxide is produced associated with 90 kg of unreacted glucose. What are the mass percentages of ethanol and propenoic acid at the output of the must? Consider that no glucose is assimilated by the bacterium.

Answer 1

the problem is based on the concept of material balances.

the mass balance states that total amount of the component coming into the system must comes out of the system.

a)

in this problem we are given that coal is burned with theoretical amount of oxygen.

the components if the coal are:

components %moles (mol/mol) с 84.36 H2 1.89 02 4.4 N2 0.63 S 0.89 Ash 7.83

let us assume that total moles of the coal coming in is 100 moles.

so based on that moles fo each component will be ;

components %moles (mol/mol) moels (mol) C 84.36 84.36 H2 1.89 1.89 02 4.4 4.4 N2 0.63 0.63 S 0.89 0.89 Ash 7.83 7.83

when the coal is burned with theoretical oxygen , C, C and H2 will react to form the product gases.

the reactions will be,

C+02 CO

so, from these reaction , we can see that 1 mol of C burns with 1 mol of O2 t o produce 1 mol of CO2.

so, 84.36 mol of C requires 84.36 mol of O2 to produce 84.36 mol.

so,

moles of O2 required = 84.36 mol

moles of CO2 produced = 84.36 mol

now for H2, reaction will be,

$H_2O + 0.5O_2\rightarrow H_2O$

so, from these reaction we can see that 1 mole of H2 reacts with 0.5 mol of O2 to get 1 mol of H2O.

hence for 1.89 mol of H2 we get,,

moles of O2 requires = 0.5 * 1.89 mol = 0.945 mol

moles of H2O produced = 1.89 mol

for S , reaction will be,

S+02 SO

so from the stoichiometry of the reaction , we can see that 1 mol of S reacts to produce 1 mole of SO2 with 1 mol of O2

so,

for 0.89 mol oS we get,

moles of O2 required = 0.89 mol

moles of SO2 produced = 0.89 mol

so based on the above calculations,

the total moles of O2 required = 0.89 mol + 84.36 mol + 0.945 mol = 86.195 mol

so, the theoretical moles fo Air fed will be,

Air fed = moles total of O2 required / 0.21 = 86.195 mol/0.21 = 410.452 mol

hence,

the O2 fed = 86.195 mol

the moles of N2 fed with air = 0.79 * air fed = 0.79 * 410.452 mol = 324.25 mol

so,

the moles of N2 coming out = moles of N2 coming in air + moles of N2 coming in with coal = 0.63 mol + 324.25 mol = 324.88 mol

so, the components and moles table for the outlet components will be,

component outlet moles (mol) CO2 84.36 H20 1.89 02 4.4 N2 324.88 SO2 0.89 Ash 7.83

and the mole fraction of the components with out ash basis will be.

total moles except ash = 416.42 mol

so, moles fraction of CO2 = moles of CO2 / total moles without ash = 84.36 mol / 416.42 mol = 0.2025 mol/mol.

similarly, doing this for all the components without ash, we get the following table,

component outlet moles (mol) mole fraction (mol/mol) CO2 84.36 0.20258393 H20 1.89 0.004538687 02 4.4 0.010566255 N2 324.88 0.780173863 SO2 0.89 0.002137265

so, the above table is the answer .

b)

we are given the reaction as,

$Sb_2S_3 + 3Fe\rightarrow 2Sb + 3FeS$

from the reaction we can see that according to the stoichiometry , 1 mol og Sb2S3 reacts with 3 moles of Fe completely to produce 2 moles of Sb.

so,

the data given to us is,

the mass of Sb2S3 fed = 0.60 kg

mass of Fe fed = 0.250 kg

we know that

molar mass of Sb2S3 = 679.43 kg/kmol

molar mass of Fe = 56 kg/kmol

so,

the moles of Sb2S3 fed = mass / molar mass = 0.60 kg/679.43 kg/kmol =

and

the moles of Fe fed = 0.250 kg/56 kg/kmol =

now the concept of limiting reagent is that the component which gets completely reacted , which means that the moles of that component will be less compared to the other reactant.

so, we can see from the stoichiometry of the reaction that, 1 mol of Sb2S3 reacts with 3 mol of Fe.

so, mol of Sb2S3 will requires , 3 * =

so, the moles of Fe required to react completely with Sb2S3 is lesser than the fed moles of Fe ,

this means that Fe is in excess.

so,

1)

the limiting reagent is Sb2S3. and excess reagent is Fe.

2)

the % excess of Fe = (moles of Fe fed - moles of Fe required) / moles of Fe required = = 0.6817

so,

the % excess Fe is 68.17 %.

3)

the extent of the reaction is given as ,

$\zeta = \frac{final \ moles - initial \ moles }{v}$

here,

v = stoichiometric coefficient (take + for the product and - for the reactant)

so,

now we have been given that,

mass of Sb produced = 0.2 kg

we know ,

molar mass of Sb = 121.76 kg/kmol

so,

moles of Sb produced = mas /molar mass = 0.2 kg/121.76 kg/kmol =

so,

the initial moles of Sb = 0 kmol

final moles of Sb =

v = +2

so,

$\zeta = \frac{final \ moles - initial \ moles }{v}$

$\zeta = \frac{1.642 * 10^{-3}kmol - 0 kmol }{2} = 0.821 * 10^{-3} kmol$

so, the extent of reaction is .

4)

let X be the conversion of Sb2S3,

so,

the moles of Sb produced = 2 * moles of Sb2S3 fed * X

so,

we get

we get

X = 0.9297 mol/mol

hence the conversion of Sb2S3 is 972.97 %.

5)

yield = mass of Sb produced / mass of Sb2S3 fed = 0.2 kg/0.6 kg = 0.333 kg/kg

so,

the yield is 0.333 kg.

c)

we are given two reactions,

Reaction 1: C6H1206+ 2C2H5OH +2CO2 Reaction 2: C6H12062C2H3CO2H +2H20

we have the data as,

total mass of solution = 3500 kg

weight fraction of glucose = 0.12 kg/kg

so,

the mass of glucose fed = 0.12 kg/kg * 3500 kg = 420 kg.

now,

mass of glucose un reacted = 90 kg

so,

mass of glucose reacted = total mass - un reacted mass = 420 kg - 90 kg = 330 kg.

we have

mass of CO2 produced = 120 kg.

we know that,

molar mass of glucose = 180 kg/kmol

molar mass of CO2 = 44 kg/kmol

molar mass of ethanol = 46.07 kg/kmol

molar mass of propinoic acid = 74 kg/kmol

molar mass of water = 18 kg/kmol

so,

the moles of glucose reacted = mass /molar mass = 330 kg / 180 kg/kmol = 1.833 kmol

and

the moles of CO2 produced = 120 kg/44 kg/kmol = 2.7272 kmol

now,

from the reaction 1, we can see that according to the stoichiometry of the reaction, 1mol of glucose produces 2 mol of CO2,

hence , to produce 2.7272 kmol of CO2 we need, 2.7272kmol/2 = 1.3636 kmol og glucose.

so, the glucose reacted in reaction 1 = 1.3636 kmol

that means that rest of the gluse will give propinoic acid,

so,

the moles of glucose reacted in reaction 2 = total reacted moles - moles reacted in reaction 1 = 1.833 kmol - 1.3636 kmol = 0.4693 kmol

so,

the moles of ethanol produced = moles of CO2 produced = 2.7272 kmol

moles of propinoic acid produced = 2*moles of glucose reacted in reaction 2 = 2 * 0.4693 kmol = 0.9387 kmol

the moles of CO2 produced = 2.7272 kmol

the moles of H2O produced = moles of propinoic acid produced = 0.9387 kmol

so,

in terms of mass,

mass of ethanol produced = moles * molar mass = 2.7272 kmol * 46.07 kg/kmol = 125.642 kg

mass of propinoic acid = 0.9387 kmol * 74 kg/kmol = 69.46 kg

mass of H2O = 16.890kg

mass of CO2 = 120kg

and mass of glucose un reacted = 90 kg

so, the total mass of outlet = 90 + 125.642 + 120+16.89+69.46 = 421.99 kg

so,

the weight % of propinoic acid = mass of propinoic acid / total mass * 100 = 69.46 kg / 421.99 kg *100 =16.46%.

the weight % of ethanol = mass of ethanol / total mass * 100 = 125.642 kg / 421.99 kg *100 =29.77%.

so,

the mass percentages of ethanol and propinoic acid at the output are 29.77 % and 16.46% respectively.