Question

1

Predict the product of the following reaction and classify the reaction. 

 Pb+ FeSO4→ PbSO4+

2 

When 5.00 g of FeCl3 xH2O are heated, 2.00 g of water are driven off. Find the chemical formula of the hydrate and the % mass H2O composition. 

Answer 1

Solution:

Question 1.

In the given reaction, Pb + FeSO₄ PbSO₄ + Fe, the oxidation state of Pb changes from 0 to +2 and Fe changes from +2 to 0. Therefore, the reactant Pb is oxidized and Fe is reduced.

Hence, the product is Fe and the reaction is redox reaction.

Question 2.

1. Given data:

Mass of FeCl₃.xH₂O = 5.00 g

Amount of water driven off = 2.00 g

Mass of FeCl₃ = (Mass of FeCl₃.xH₂O - Amount of water) = 5.00 g - 2.00 g = 3.00 g

2. Moles of FeCl₃ & H₂O:

Moles (n) = Mass / Molar mass

n_FeCl3 = 3.00 g / (162.2 g/mol) = 0.018496 mol

n_H2O = 2.00 g / (18.01528 g/mol) = 0.11102 mol

3. Ratio of FeCl₃ & H₂O:

Divide the individual mole by the least mole. Therefore,

FeCl₃ : H₂O = (0.018496 mol / 0.018496 mol) : (0.11102 mol / 0.018496 mol) = 1 : 6

Thus, the chemical formula of the hydrate is FeCl₃.6H₂O.   

4. % mass H₂O:

Mass of H₂O in FeCl₃.6H₂O = 6 x 18.01528 g/mol = 108.09168 g/mol

Molar mass of FeCl₃.6H₂O = 270.29 g/mol

% Mass H₂O = (Mass of H₂O in FeCl₃.6H₂O / Molar mass of FeCl₃.6H₂O) x 100%

= (108.09168 g/mol / 270.29 g/mol) x 100% = 39.99 % = 40%

Hence, 40% is the % mass H₂O composition.