n=20 mol
(NH)4 SO4
Atomic masses :
N- 14
H- 1
S- 32
O- 16
Therefore M= 14×2 + 1×8 + 32 + 16×4
= 132
m= nM
= 20×132
= 2640g
<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
------------------------------
1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
If its one part magnesium and two parts hydroxide id say 88g of magnesium
Answer:
The conversion achieved for the first CSTR impeller is 0.382
Discrepancy = 0.188
Explanation:
The impeller divides the CSTR into 2 equal reactors of volume 500gal
Using V = FaoX/ (-ra)
500gal = Fao×Xa/[(KCao^2( 1 -X1)^2]
500gal = CaoVoX1/ KCao^2(1-X1)
500gal= 500gal × X1'/(1 - X1)^2
(1 -X1)^2 = X1
X1^2 - 3X1 + 1 = 0
X1= 0.382
Conversion achieved in the first CSTR is 0.382
Actual measured CSTR = 57% =57/100=0.57
Discrepancy in the conversions= 0.57 -0.383 =0.188
