Answer to Question #83274 in Chemistry for Divine

Question

How many aluminium ions will be discharged by 0.33F?[Avogadro constant = 6.02*10^23

Answer 1

.n (Al^(3+)) =?

.m (F^-) =0.33

N_A= 6.02× 10^23

Find the amount of substance of F^-

.n=m/A_r = 0.33/19 =0.017 mol

Reaction between Al^(3+) and F^- is

Al^(3+) +3F^-= AlF_3

. n (Al^(3+))= 1/3 n (F^-)= 0.017/3 =0.0057

. n (ions of Al^(3+))= n× N_A=0.0057×6.02×10^23 =3.43×10^21