Oblicz, ile gramów wodorotlenku miedzi(II) wytràciło si´ w wyniku dodania 270g 25-procentowego roztworu chlorku miedzi(II) do 400 g 10-procentowego roztworu wodorotlenku sodu. potrzebne na teraz
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CuCl2:
mr=270g
Cp=25%
ms=Cp*mr/100%
ms=25*270/100
ms = 67,5g
NaOH:
mr=400g
Cp=10%
ms=Cp*mr/100%
ms=10*400/100
ms = 40g
mNaOH=40u
mCuCl2=135u
mCu(OH)2=98u
2NaOH + CuCl2 ---> Cu(OH)2 + 2NaCl
80g NaOH -----135g CuCl2
40g NaOH ----- xg CuCl2
x = 67,5g CuCl2
80g NaOH ----- 98g Cu(OH)2
40g NaOH ------ xg Cu(OH)2
x = 49g Cu(OH)2