D. Sodium hydroxide aka naOH
Answer:
(a) The proportion of dry air bypassing the unit is 14.3%.
(b) The mass of water removed is 1.2 kg per 100 kg of dry air.
Explanation:
We can express the proportion of air that goes trough the air conditioning unit as
and the proportion of air that is by-passed as
, being
.
The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

Replacing the first equation in the second one we have

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.
It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.
The water removed of every 100 kg of dry air is

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).
100 * 0.012 = 1.2 kgW
PV = nRT
Where:
P = pressure in atm = 700/760 = 0.9211atm
V = volume = 8.29L
R = gas constant, 0.08206 atm-L/mol-K
T = temperature in Kelvin = 200 + 273 = 473
n = numbers of moles = Mass/molar mass
mass of the compound = 30.5
we can rewrite the equation above as
PV = (Mass)/(Molar mass) * RT
The proper name for PCI6 is phosphorus hexachloride. Therefore it’s true