+nAl = 3,24/27 = 0,12 mol
PT
2Al + 6HCl -> 2AlCl3 + 3H2
0,12_0,36____0,12_____0,18(mol)
VH2 = 0,18*22,4 = 4,032 lít
mH2 = 0,18 *2 = 0,36g
mHCl (dd HCl) = 0,36 * 36,5= 13,14 g
-> mdd HCl cần dùng = 13,14 / 20% = 65,7g
mAlCl3 = 0,12 * 133,5 = 16,02g
m dd AlCl3 = mAl+mddHCl-mH2 = 3,24+65,7-0,36 = 68,58g
-> C%dd AlCl3 = 16,02/68,58 *100%= 23,36%
n\(_{Al}\)= \(\dfrac{3,24}{27}\)= 0,12 (mol)
PTHH: 2Al + 3H2SO4 ----> Al2(SO4)3 + 3H2\(\uparrow\)
mol: 0,12------>0,18------------>0,12------>0,18
V\(_{H_2}\)= 0,18 . 22,4 = 4,032 (mol)
m\(_{H2SO4}\)= 0,18 . 98 = 17,64 (g)
mdd\(_{H_2SO_4}\)= \(\dfrac{17,64}{20}\).100 = 88,2 (g)
m\(_{Al_2\left(SO_4\right)_3}\)= 0,12 . 342 = 41,04 (g)
mdd sau pứ = 3,24 + 88,2 = 91,44 (g)
C% Al\(_2\)(SO4)\(_3\) = \(\dfrac{41,04}{91.44}\).100% = 44,88%