Răspuns:
Mg+O2=2MgO
MgO+H2O=Mg(OH)2
5KMOLI =5×1000=5000 moli Mg(OH)2
m=5×MMg(OH)2
MMg(OH)2=24+2(16+1)=24+34=78g/mol
5Kmoli=5×78= 390 Kg hidroxid de magneziu
MgO+ H2O = Mg(OH)2
MMgO=24+16=40g/mol
40g................78g
X.....................390Kg Mg(OH)2
X=(40×390)÷78=200Kg MgO oxid de
de magneziu
Mg+O2 =2MgO
24g Mg..........2×40gMgO
X........................200KgMgO
X=(24×200)÷80=4800÷80=60KgMg
puritatea =(masa pură×100)÷masa impură
p=(mp×100)÷ mi
mp=(p×mi)÷ 100
mp=(90×60)÷100=4500÷100=45KgMg
