Question

Determina cantitatea de clorura de aluminiu in moli obtinuta in reactia aluminiului cu 15 mili de clor ​

Answer 1

Răspuns:

MCl2=2×35,5=71g/mol

n=m/M

m=n×M=115×71=1.065g

MAlCl3=27+3×35,5=133,5g/mol

3×71 2× 133,5g

2Al + 3Cl2 => 2AlCl3

1.065g xg

x=1.065×2×133,5/3×71= calculează