A 25.9 g nickel plate was immersed in 555 g of an iron (III) sulfate solution with a salt fraction of 0.1.

A 25.9 g nickel plate was immersed in 555 g of an iron (III) sulfate solution with a salt fraction of 0.1. After keeping the plate in solution for some time, it was taken out, and it turned out that the mass fraction of iron (III) sulfate became equal to the mass fraction of the formed nickel (II) salt. Determine the mass of the plate after removing it from the solution.

Given:
m out. (plates) = 25.9 g
m out. solution (Fe2 (SO4) 3) = 555 g
ω ref. (Fe2 (SO4) 3) = 0.1
ω end. (Fe2 (SO4) 3) = ω (NiSO4)

Find:
m end. (plates) -?

1) Ni + Fe2 (SO4) 3 => NiSO4 + 2FeSO4;
2) If ω end. (Fe2 (SO4) 3) = ω (NiSO4), so m end. (Fe2 (SO4) 3) = m (NiSO4);
3) Let n react. (NiSO4) = (x) mol;
4) m react. (NiSO4) = n reag. (NiSO4) * M (NiSO4) = x * 155 = (155x) g;
5) n react. (Fe2 (SO4) 3) = n reag. (NiSO4) = (x) mol;
6) m react. (Fe2 (SO4) 3) = n reag. (Fe2 (SO4) 3) * M (Fe2 (SO4) 3) = x * 400 = (400x) g;
7) m end. (Fe2 (SO4) 3) = m ref. (Fe2 (SO4) 3) – m reag. (Fe2 (SO4) 3);
8) m out. (Fe2 (SO4) 3) = ω ref. (Fe2 (SO4) 3) * m ref. solution (Fe2 (SO4) 3) = 0.1 * 555 = 55.5 g;
9) 155x = 55.5 – 400x;
x = 0.1;
10) n react. (Ni) = n react. (NiSO4) = x = 0.1 mol;
11) m reag. (Ni) = n react. (Ni) * M (Ni) = 0.1 * 59 = 5.9 g;
12) m end. (plates) = m ref. (plates) – m react. (Ni) = 25.9 – 5.9 = 20 g.

Answer: The mass of the plate is 20 g.