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K₂O + H₂O -----> 2KOH
94u 112u
m K₂O = 2 * 39u + 16u = 94u
m 2KOH = 2 * 39u + 2 * 16u + 2 * 1u = 78u + 32u + 2u = 112u
układamy proporcję
94u K₂O --------------- 112u KOH
47g K₂O --------------- x
x = 47g * 112u / 94u
x = 56g
Z 47g tlenku potasu powstanie 56g wodorotlenku potasu
Z równania reakcji : M K₂O = 94 g/mol-------------- 2* M KOH=2* 56g= 112g/mol
--------------------------------m K₂O = 47g --------------------- m KOH = x
x= 47* 112 / 94
x= 56 [g]