Question

4.08 g of a mixture of BaO and unknown carbonate MCO₃ was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of I NHCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M.

Answer 1

Given that : that weight of the mixture of BaO and MCO₃

= 4.08 g

and the weight of the residue = 3.64 g

:. The weight of CO₂ evolved

 = 4.08 - 3.64 = 0.44 g

Thus the number of moles of CO₂ evolved

= 0.44/44 = 0/01

Hence, the number of moles of MCO₃ ≡  0.01 Volume of 1 NHCI which dissolved the residue = 0.01 mL

The volume of 1 NHCl which was required for dissolution of the residue

 

The dissolution of residue (BaO + MO) is given as :

BaO + MO + 4HCl → BaCl₂ + MCl₂ + 2H₂O 

Where, 2 moles of residue (BaO + MO)are dissolved. Thus, the number of moles of residue

(BaO + MO) = 0.06/2 = 0.03

Number of moles of MO = No. of moles of MCO₃ = 0.01

.'. Number of moles of BaO = 0.03 - 0.01 = 0.02 

Weight of I mole of BaO : 154

('.' molecular wt of BaO = 154)

'. Weight of 0.02 mole of BaO = 154 x 0.029

= 3.08 g

:. Weight of 0.01 mole of MCO₃ 

= weight of the mixture - weight of BaO 

= 4.08- 3.08 = 1.00 g

Since the atomic weight of calcium (Ca) is 40, therefore, the metal 'M' is calcium.