What is the mass of the precipitate formed by the interaction of an excess of a zinc chloride solution weighing

What is the mass of the precipitate formed by the interaction of an excess of a zinc chloride solution weighing 160 g with a mass fraction of a substance of 15%, with sodium hydroxide?

Given:
m solution (ZnCl2) = 160 g
ω (ZnCl2) = 15%

Find:
m (draft) -?

Solution:
1) ZnCl2 + 2NaOH => Zn (OH) 2 ↓ + 2NaCl;
2) M (ZnCl2) = Mr (ZnCl2) = Ar (Zn) + Ar (Cl) * 2 = 65 + 35.5 * 2 = 136 g / mol;
M (Zn (OH) 2) = Mr (Zn (OH) 2) = Ar (Zn) + Ar (O) * 2 + Ar (H) * 2 = 65 + 16 * 2 + 1 * 2 = 99 g / mole;
3) m (ZnCl2) = ω (ZnCl2) * m solution (ZnCl2) / 100% = 15% * 160/100% = 24 g;
4) n (ZnCl2) = m (ZnCl2) / M (ZnCl2) = 24/136 = 0.18 mol;
5) n (Zn (OH) 2) = n (ZnCl2) = 0.18 mol;
6) m (Zn (OH) 2) = n (Zn (OH) 2) * M (Zn (OH) 2) = 0.18 * 99 = 17.82 g.

Answer: The mass of Zn (OH) 2 is 17.82 g.




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