Răspuns:
Explicație:
1.
2Na +2H2O= 2NaOH +H2
Na2O + H2O=2NaOH
2H2O + Ca= Ca(OH)2 +H2
CaO + H2O=Ca(OH)2
2NaOH + CuSO4=Na2SO4 + Cu(OH)2
2.
Ca(OH)2 +CO2=CaCO3 +H2O
3.
NaOH
4.
xg 19,6g
2NaOH + CuSO4= Cu(OH)2 +Na2SO4
160g 98g
x=160.19,6 : 98=32g CuSO4
n=32g :160g/moli=0,2moli CuSO4
5.
5,6g xg zg yg
2K + 2H2O=2KOH +H2
2 .39g 2.18g 2.56g 2g
-se afla care dintre reactanti este in exces
x=5,6.36 : 78g= 2,58 g apa care se consuma
Deci apa este in exces-----> 120 - 2,58=117,42 apa in exces
-se afla VH2
y=5,6.2 : 78=0,14 g H
n=0,14 g : 2g/moli=0,07moli H2
V= 0,07 moli.22,4 L/moli =1,368L
-se afla masa de KOH = z
z=5,6. 2 .56 :78=8,04g KOH
-se afla masa sol. de KOH
ms= md + mapa ramasa
ms = 8,04 +117,42=125,46g
-se afl concentratia sol. finale
c= md.100 : ms
c=8,04 .100 : 125,46=6,4 %
verifica calculele!
