Dau coroana si 50 de puncte, repedee va rog multt
Răspunsuri la întrebare
Răspuns:
Explicație:
1.
160g NaOH
masa de HCl
160g xg
NaOH + HCl= NaCl + H2O
40g 36,5g
x= 160 . 36,5 : 40= 146g HCl
MNaOH=23 + 16 + 1=40----> 50g/moli
MHCl= 1+ 35,5=36,5----> 36,5g/moli
2.
27 g Al
react. cu 32 g O2
masa de Al2 O3
27g xg yg
4Al + 3 O2 = 2Al2O3
4 .27g 3 .32g 2. 102g
x= 27 . 3 .32 : 4 . 27 =24 g O2 consumat in reactie
Deci oxigenul este in exces -----> 32g - 24g -8g O in exces
y= 27 . 204 : 108= 51 g Al2O3
3.
20g CuSO4
react. cu NaOH
a. moli NaOH
b. masa de Na2SO4
xg 20g yg
2NaOH + CuSO4= Na2SO4 + Cu(OH)2
2.40g 160g 142g
x= 20 . 80 : 160=10 g NaOH
n= m : M
n= 10g : 40g/moli= 0,25 moli NaOH
y= 20 . 142 : 160=17,75 g Na2SO4
MNaOH= 23 + 16 + 1=40------> 40g/moli
MCuSO4= 64 + 32 + 4.16= 160----> 160g/moli
MNa2SO4= 2.23 + 32 + 4.16= 142------> 142g/moli