A biofuel researcher is running an experiment to produce methane from wastewater treatment plant sludge. The researcher places the sludge and a microbial inoculum into a sealed glass bottle rated to 20 psi, and is worried that if the pressure within the bottle becomes too high, that the glass may break. During the experiment, the aqueous concentration of methane in the closed bottle reaches 20 ppm. Determine the partial pressure of methane (PCH4) in the headspace at equilibrium if the KH, CH4

Missing data related to question is attached below

If the KH,CH4 = 1.42 * 10^-3 mol/L atm

0.88 atm

Explanation:

P1 = 20 psi

aqueous concentration of methane in closed bottle = 20 ppm

20 ppm = 20 / 1000 ( gram / Kg ) =  0.020 gram/kg = 0.020 g/liter

molar mass of methane = 16 gram/mol

next : convert mass conc to molar conc

Cm  = 0.020 / 16

Cm = 1.25 * 10^-3 mol/L

Given that KH,CH4 = 1.42 * 10-3 mol/L atm

Applying equilibrium relationship

Cm = ( KH,CH4 ) ( partial pressure of methane )

hence partial pressure of methane

= Cm / ( KH,CH4 )  = 1.25 * 10^3 / (1.42 * 10^-3 ) = 0.88 atm

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"