How many grams of potassium zincate (K2ZnO2) are formed if 2 moles of zinc hydroxide

| March 12, 2021 | education

How many grams of potassium zincate (K2ZnO2) are formed if 2 moles of zinc hydroxide are exposed to excess potassium hydroxide?

Given:

n (Zn (OH) 2) = 2 mol

m (K2ZnO2) -?

Decision:

Zn (OH) 2 + 2KOH → K2ZnO2 + 2H2O

1. Determine the molecular weights of zinc hydroxide and potassium zincate:

M (Zn (OH) 2) = 65 + 16 * 2 + 1 * 2 = 99 g / mol;

M (K2ZnO2) = 39 * 2 + 65 + 16 * 2 = 175 g / mol.

2. Calculate the masses of zinc hydroxide and potassium zincate according to the reaction equation:

m (Zn (OH) 2) = n * M = 1 mol * 99 g / mol = 99 g;

m (K2ZnO2) = n * M = 1 mol * 175 g / mol = 175 g.

3. Calculate the mass of zinc hydroxide reacted:

m (Zn (OH) 2) = n * M = 2 mol * 99 g / mol = 198 g.

We calculate the mass of zinc hydroxide according to the proportion:

99 g —- 175 g

198 g —– x, where x is the mass of potassium zincate reacted.

x = (198 g * 175 g) / 99g = 350 g.

Answer: 350 g of potassium zincate.



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