Answer to Question #289864 in Chemistry for Peace

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