Question 1:(10 Marks) Stoichiometric equations can be used to represent the grow

Question

Question 1:(10 Marks) Stoichiometric equations can be used to represent the growth of microorganisms provided that a "inolecular formula" for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, O, and other elements in the cells. For a particular bacterial strain, the molecular formula was determined to be CisH oNo.m. The bacteria are grown under aerobic conditions with hexadecane (C.sH) as a substrate. The reaction equation describing growth is as follows a) Is this stoichiometric equation balanced? If not, how would you balance this equation? (2 Mark) b) Assuming a 100% conversion rate, what is the yield of cells from hexadecane in g/g? (4 Marks) c) Assuming a 100% conversion rate, what is the yield of cells from oxygen in g/g? (4 Marks) (Hint: MM= 16g/g-mol; MM C-12g/g-mol; MM N #14g/g-mol; MM H = 18/g-mol)

Explanation / Answer

Answer a)

First of all the equation is balanced or not we have to observe

The equation is

C16H34 + 1.42NH3 + O2 => 1.65 C4.4H7.3O1.2N0.86 + 8.74CO2 + 13.11H2O

Number of C in left side = 16

Number of C in right side = (1.65*4.4+8.74*1) = 16; so C atom is balanced

H in left side = (34 + 1.42*3)= 38.3

H in right side = (1.65* 7.3+13.11*2)=38.3 ; so H atom is balanced

N in Left side = 1.42

N in Right side = (1.65*0.86)=1.42 ; so N is balanced

O in left side =2

O in right side = (1.65*1.2+8.74*2+13.11) = 32.6 ; so we have to multiply O by 16.3

So Balanced equation is

C16H34 + 1.42NH3 + 16.3 O2 = 1.65 C4.4H7.3O1.2N0.86 + 8.74CO2 + 13.11H2O

Answer b)

So for calculation of yield,

So from the equation 1 mole of hexadecane is giving 1.65 mole of biomass.

Now 1 mole of Hexadecane = (16* 12 + 34*1) = 226 g of Hexadecane

Now 1 mole of product biomass = (4.4*12 +7.3*1+1.2*16+0.86*14)= 91.34 g of product

Now 1.65 mole of product = 91.34*1.65 = 150.7 g.

So Percent of yield = 150.7/226 % = 66.7 %

Answer c)

Form the previous calculation,

1.65 mole of product = 91.34*1.65 = 150.7 g.

1 moles of O2 is equals to 16*2 = 32 g of O

Now 16.3 moles of O2 = 32*16.3 = 521.6 g of O

Now yield from oxygen = 150.7/521.6% = 28.9 %

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