Assume the Henry’s coefficient for water at $\pu{20 °C}$ is $\pu{1.6E8 Pa}.$ What is the mole fraction of $\ce{CO2}$ that will be dissolved in water assuming that it is at equilibrium with a flue gas containing $0.11$ mole fraction $\ce{CO2}$ at atmospheric pressure $(\pu{10^5 Pa})?$

My understanding is that we can use Henry's law

$$y_i\cdot p = x_i\cdot H$$

where $H$ is the given Henry's coefficient. This yields a solution for $x(\ce{CO2}) = \pu{6.89E-5}.$

A coal plant emits $\pu{120 kg s-1}$ of $\ce{CO2}$ at a mole fraction of $0.11$ and atmospheric pressure $(\pu{10^5 Pa}).$ Assuming that the flue gas and water come fully into equilibrium during separation, what flow rate of water $[\pu{m3 s-1}]$ would be needed to capture $95\%$ of the $\ce{CO2}?$

For this part, I'm a bit confused as to where our $x(\ce{CO2})$ and the $95\%$ capture rate comes into the play. I assume after you calculate a new mole fraction, some dimensional analysis is involved to get from $\pu{kg s-1}$ to $\pu{m3 s-1}.$

Any insight to this would be appreciated.

  • $\begingroup$ I guess you must have heard about concentration kg/m3 and equation for ideal gas density=f(M,p,T) $\endgroup$ – Poutnik Oct 22 at 6:33

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