Question

In reactia dintre sulfat de zinc si hidroxid de potasiu se utilizeaza 2300g solutie hidroxid de potasiu c=20%. Sa se calculeze nr. de moli de sulfat de zinc si nr. de moli(si cantitate) de produsi de reactie.

Answer 1

ZnSO4 + 2 KOH = Zn(OH)2 + K2SO4
c = md/ms * 100
md = 2300*20/100 = 460 g KOH
M KOH = 39+16+1 = 56
M Zn(OH)2 = 65+2*(16+1) = 99
M K2SO4 = 39*2+32+4*16 = 174
1 mol ZnSO4..............2*56 g KOH........99 g Zn(OH)2..........174 g K2SO4
x.................................460 g KOH.....................y..............................z
x = 4,107 moli ZnSO4
y = 406,607 g Zn(OH)2
n = 406,607/99 = 4,107 moli Zn(OH)2
z = 714,642 g K2SO4
n = 714,642/174 = 4,107 moli K2SO4