x-Al
y-Al2O3
x + y = 10g
2 moli...........................3*2=6g
2Al + 6HCl => 2AlCl3 + 3H2
x moli ..............................0,6g
x = 0,2 moli Al
m Al = 0,2*27 = 5,4 g
5,4 + y = 10 => y = 4,6 g (Al2O3)
Al2O3 + 6HCl => 2AlCl3 + 3H2O
%Al = mAl/m amestec*100 = 5,4/10*100 = 54
%Al2O3 = 46
6*36,5g HCl....................3*2 = 6g H2
xg HCl.............................0,6 g
x= 21,9g HCl
102g Al2O3........................6*36,5g HCl
4,6g Al2O3............................x = 9,876g HCl
md total = 9,876+21,9 = 31,77g
C = md/ms*100 => ms = md*100/C = 3177/17,3 = 183,679g sol. HCl