Answer:107.1 g, 124.1 g
Explanation:
The equation of the reaction is;
Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)
Hence;
For Al2S3
Number of moles= reacting mass/molar mass
Number of moles = 158g/150gmol-1 =1.05 moles
If 1 mole of Al2S3 yields 3 moles of H2S
1.05 moles of Al2S will yield
1.05 × 3/1 = 3.15 moles
Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g
For water
Number of moles of water = 131g/18gmol-1= 7.3 moles
6 moles of water yields 3 moles of H2S
7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S
3.65 moles × 34 gmol-1 =124.1 g
Answer:
As temperature increases the volume of given amount of gas increases while pressure and number of moles remain constant.
Explanation:
According to the charle's law,
The volume of given amount of gas is directly proportional to the temperature at constant pressure and number of moles of gas.
Mathematical expression:
V ∝ T
V = KT
V/T = K
When temperature changes from T₁ to T₂ and volume changes from V₁ to V₂.
V₁/T₁ = K V₂/T₂ = K
or
V₁/T₁ = V₂/T₂
Thus, the ratio of volume and temperature remain constant for constant amount of gas at constant pressure.
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
