Answer to Question #98963 in Microbiology for amir

Question

A 10 g (dry weight) mouse is accidentally drowned in a 25 L bottle of non-sterile
spring water that initially contains 10 mg/L oxygen. Assume that 70% of the
mouse’s dry weight is CH2O.
a. Write the relevant reaction for respiration of the mouse with oxygen.
b. Does the jug become anaerobic?
c. If 2 ppm NO3- and 15 ppm SO42- are in the spring water, how much of each
will be left when the jug reaches equilibrium? Write relevant reactions.
d. How much of the mouse will be left, assuming no organism capable of
fermentation are present?

Answer 1

a)$CH_2O+O_2->CO_2+H_2O$

b)No of moles of $O_2=(10×25×10^{-3})/32=0.0078moles$

No of moles of $CH2O =10/30.03=0.233moles$

$CH_2O+O_2->CO_2+H_2O$

0.233mole 0.0078moles

No of moles of CH2O $left= 0.233-0.0078=0.2252$

No of moles of Oxygen is less.Oxygen is limiting reagent.So,the jug becomes anaerobic.

c)$2CH_2O + SO_4^{2- }+ 2H^+ -> H_2S + 2CO_2 + 2H_2O$

0.2252 15×0.2252/10⁶

0.2252-2×3.37×10^-6 0

0.2252 0

$4NO_3^-+ 5CH_2O ->2N_2O + 5CO_2 + 5H_2O$

0.2252 2×0.2252/10⁶

0.2252-4.5×4×10^-7/5 0

0.2252moles

At equilibrium,$NO_3^-$ and $SO_4^{2-}$ are completely exhausted.

d)$mass=0.2252×30.03=6.763g$ of $CH_2O$

Total mass of mouse $left =(3+6.763)g=9.763g$ of dry weight of mouse is left.