Carbon Capture Absorption with Henry's Law


> 
> 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.


