a)
5,4 g md m g
2Al + 6HCl --> 2AlCl3 + 3H2
2x27 6x36,5 2x187,5
=> md = 5,4x6x36,5/2x27
= 21,9 g HCl
din c% = mdx100/ms
=> ms = mdx100/c%
= 21,9x100/20 = 1095 g sol HCl
m.apa = ms-md = 1095-21,9
= 1073,1 g
=> m = 5,4x2x187,5/2x27
= 37,5 g AlCl3
c%,dupa reactie = mx100/m.apa+m
=> c% = 37,5x100/(1073,1+37,5)
= 3750/1110,6 = 3,37%
b)
5,6 g md m g
Fe + 2HCl --> FeCl2 + H2
56 2x36,5 112
=> md = 5,6x2x36,5/56 = 7,3 g HCl
din c = mdx100/ms
=> ms = mdx100/c
= 7,3x100/10 = 730 g sol HCl
=> m = 5,6x112/56 = 11,2 g FeCl2
m.apa = ms-md = 730-7,3 = 722,7 g
=> ms = 722,7+11,2 = 733,9 g sol. FeCl2
=> c% = 11,2x100/733,9 = 1,53%