BaCl2 + K2SO4 = BaSo4 + 2KCl
200 гр ----- 100%
Х ----- 41.6 %
Х= 83.2 m (BaCl2)
400 --- 100%
X ---- 17.4%
X= 69.6 m (K2SO4)
n(BaCl2) = 83.2/ 208 = 0.4
m (BaSO4) = 0.4• 233= 93.2 гр.
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BaCl2 + K2SO4 = BaSo4 + 2KCl
200 гр ----- 100%
Х ----- 41.6 %
Х= 83.2 m (BaCl2)
400 --- 100%
X ---- 17.4%
X= 69.6 m (K2SO4)
n(BaCl2) = 83.2/ 208 = 0.4
m (BaSO4) = 0.4• 233= 93.2 гр.