The zinc plate was placed in 383.4 g of a solution of ferrous sulfate 3 after a while the plate

The zinc plate was placed in 383.4 g of a solution of ferrous sulfate 3 after a while the plate was taken out, dried and weighed. it became lighter by 16.6 g. Calculate the mass of zinc passed into the solution, the mass of iron released on the plate, and the mass fraction of zinc sulfate formed in the solution

Given:
m solution (Fe2 (SO4) 3) = 383.4 g
Δm (plates) = 16.6 g

To find:
m react. (Zn) -?
m (Fe) -?
ω (ZnSO4) -?

Decision:
1) 3Zn + Fe2 (SO4) 3 => 3ZnSO4 + 2Fe;
2) Let n react. (Zn) = (x) mol;
5) n (Fe) = n reag. (Zn) * 2/3 = (x * 2/3) mol;
6) m react. (Zn) = n reag. (Zn) * M (Zn) = (x * 65) g;
7) m (Fe) = n (Fe) * M (Fe) = ((x * 2/3) * 56) = (112x / 3) g;
8) Δm (plates) = m react. (Zn) – m (Fe);
16.6 = 65x – (112x / 3);
x = 0.6;
n react. (Zn) = 0.6 mol;
9) m react. (Zn) = x * 65 = 0.6 * 65 = 39 g;
11) m (Fe) = 112x / 3 = 112 * 0.6 / 3 = 22.4 g;
12) n (ZnSO4) = n reag. (Zn) = 0.6 mol;
13) m (ZnSO4) = n (ZnSO4) * M (ZnSO4) = 0.6 * 161 = 96.6 g;
14) m solution = m solution (Fe2 (SO4) 3) + Δm (plates) = 383.4 + 16.6 = 400 g;
15) ω (ZnSO4) = m (ZnSO4) * 100% / m solution = 96.6 * 100% / 400 = 24.15%.

Answer: The mass of Zn is 39 g; Fe – 22.4 g; mass fraction of ZnSO4 – 24.15%.