# 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.

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