Question

При прокаливании до постоянной массы 36,6 кристаллогидрата хлорида бария получено 31,2 г твёрдого остатка. определите формулу кристаллогидрата

Answer 1

Вычислим процентное содержание воды в соединении ω(Н2О)=31,2х100/36,6=85,2%  100-85,2=14,8% воды
Мг(BaCl2)=137+71=208 формула кристаллогидрата BaCl2xnH2O m(H2O)=208x14,8/85,2=36 n=36/(2+16)=2
 BaCl2x2H2O

Answer 2

Calculating the Formula of the Hydrate

To determine the formula of the hydrate, we can use the information provided. The mass of the anhydrous salt (BaCl2) is 36.6 g, and the mass of the solid residue (hydrate minus water) is 31.2 g.

First, we need to find the mass of water lost during the heating process: - Mass of water = Mass of hydrate - Mass of solid residue - Mass of water = 36.6 g - 31.2 g - Mass of water = 5.4 g

Now, we can calculate the moles of water and the moles of the anhydrous salt to determine the formula of the hydrate.

Calculating Moles of Water and Anhydrous Salt

The molar mass of water (H2O) is 18 g/mol, and the molar mass of BaCl2 is 208.23 g/mol.

Moles of water: - Moles of water = Mass of water / Molar mass of water - Moles of water = 5.4 g / 18 g/mol - Moles of water = 0.3 mol

Moles of BaCl2: - Moles of BaCl2 = Mass of BaCl2 / Molar mass of BaCl2 - Moles of BaCl2 = 31.2 g / 208.23 g/mol - Moles of BaCl2 = 0.15 mol

Determining the Ratio of Moles

To find the ratio of moles of water to moles of BaCl2, we divide both by the smaller number of moles (0.15 mol in this case).

The ratio is approximately: - Moles of water : Moles of BaCl2 ≈ 2 : 1

Conclusion

The formula of the hydrate is BaCl2·2H2O. This means that for every mole of BaCl2, there are 2 moles of water in the hydrate.

This information is based on the principles of stoichiometry and the molar ratios of the compounds involved.