-calculez moli gaz
n=V/Vm= 13,44 l/22,4 l/mol= 0,6mol
-notez amestecul format din a molAl si b moliCu
alcatuiesc o ecuatie
ax27gAl + bx64g Al=20g
Cu nu reactioneaza cu acidul
1mol 3mol 1mol 1,5mol
Al + 3HCl -----> Al CL3 + 3/2H2 ori a moli
x...................................0,6
1,5xa=0,6molH2---> a=0,4mol Al------> m=nxM= 0,4molx27g/mol= 10,8gAl
m,Cu= m,amestec-mAl=9,2gCu
20g amestec......10,8gAl.......9,2gCu
100g.................x....................y.............................calculeaza !!!
din ecuatie, rezulta x=nHCl=1,2molHCl
m= nxM= 1,2molx36,5g/mol=43,8g---> aceasta este masa dizolvata in solutie,md
c= mdx100/ms---> ms= mdx100/c=43,8x100/30=146g solutie
