Mixed solutions containing 16.1 g of zinc sulfate and 12 g of sodium hydroxide.

Mixed solutions containing 16.1 g of zinc sulfate and 12 g of sodium hydroxide. How many and what substances will be in the solution? What substances and how much will be in the sediment?

Given:
m (ZnSO4) = 16.1 g
m (NaOH) = 12 g

Find:
m (in-in in solution) -?
m (in-in sediment) -?

Solution:
1) ZnSO4 + 2NaOH => Na2SO4 + Zn (OH) 2 ↓;
2) n (ZnSO4) = m / M = 16.1 / 161 = 0.1 mol;
3) n (NaOH) = m / M = 12/40 = 0.3 mol;
4) Zn (OH) 2 + 2NaOH => Na2 [Zn (OH) 4];
5) n (Na2SO4) = n (ZnSO4) = 0.1 mol;
6) m (Na2SO4) = n * M = 0.1 * 142 = 14.2 g;
7) n (Zn (OH) 2) = n (ZnSO4) = 0.1 mol;
8) n react. (NaOH) = n (ZnSO4) * 2 = 0.1 * 2 = 0.2 mol;
9) n rest. (NaOH) = n – n re. = 0.3 – 0.2 = 0.1 mol;
10) n (Na2 [Zn (OH) 4]) = n rest. (NaOH) / 2 = 0.1 / 2 = 0.05 mol;
11) m (Na2 [Zn (OH) 4]) = n * M = 0.05 * 179 = 8.95 g;
12) n react. (Zn (OH) 2) = n rest. (NaOH) / 2 = 0.1 / 2 = 0.05 mol;
13) n rest. (Zn (OH) 2) = n – n reag. = 0.1 – 0.05 = 0.05 mol;
14) m rest. (Zn (OH) 2) = n rest. * M = 0.05 * 99 = 4.95 g.

Answer: The mass of Na2SO4 is 14.2 g; Na2 [Zn (OH) 4] 8.95 g; Zn (OH) 2 – 4.95 g.