To a solution weighing 200 g with a mass fraction of zinc chloride of 34% was added 116 cm ^ 3 of a solution with a mass fraction

| November 17, 2020 | education

To a solution weighing 200 g with a mass fraction of zinc chloride of 34% was added 116 cm ^ 3 of a solution with a mass fraction of sodium hydroxide 32% (p = 1.35 g / cm ^ 3). Determine the mass of the sediment and the mass fractions of salts in the solution

We write down the reaction equation.
ZnCl2 + 2NaOH = 2NaCl + Zn (OH) 2.
Find the mass of zinc chloride.
m (ZnCl2) = 200 * 0.34 = 68 g.
Find the mass of sodium hydroxide.
m = pV.
m = 1.35 * 116 = 156.6 g.
m (NaOH) = 156.6 * 0.32 = 50.112 g.
Find the amount of zinc chloride substance and the amount of sodium hydroxide substance.
M (ZnCl2) = 136.3 g / mol.
M (NaOH) = 40 g / mol.
n = m / M.
n (ZnCl2) = 68: 136.3 = 0.5 mol.
n (NaOH) = 50.112 / 40 = 1.2528 mol.
Since sodium hydroxide is in excess, it means that we count by zinc chloride.
Find the mass of zinc hydroxide.
0.5 mol ZnCl2 – x mol Zn (OH) 2.
1 mol ZnCl2 – 1 mol Zn (OH) 2.
X = 0.5 * 1: 1 = 0.5 mol Zn (OH) 2.
M (Zn (OH) 2) = 99.38 g / mol.
m = 0.5 * 99.38 = 49.69 g.
Answer: m (Zn (OH) 2) = 49.69 g.



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