A 0.8027 g sample of impure al2(CO3)3 decomposed with HCL: the liberated CO2 was collected on...

A 0.8027 g sample of impure al2(CO3)3 decomposed with HCL: the liberated CO2 was collected on calcium oxide and found to weight 0.0507g calculate the percentage of aluminum in the sample

Expert Solution

The balanced equation for the given reaction is -

Al₂(CO₃)₃ (aq) + 6 HCl (aq) ----------> 2AlCl₃ (aq) + 3CO₂ (g) + 3 H₂O (l)

Mass of CO₂ produced = 0.0507 g

Therefore - Number of mols of Al₂(CO₃)₃ must present to evolve this much of CO₂ is

= (0.0507 g)(1 mol CO₂/44 g/mol)(1 mol Al₂(CO₃)₃ /3 mol CO₂)

= 3.84*10^-4 mol Al₂(CO₃)₃

Therefore, the number of mols of Al₂(CO₃)₃ in the impure sample

= [3.84*10^-4 mol Al₂(CO₃)₃]* (2 mol Al / Al₂(CO₃)₃)

= 7.68*10^-4 mol Al

Therefore - Mass of Al in the impure sample = (7.68*10^-4 mol Al)(27 g / 1 mol Al)

= 0.02074 g

Therefore, % weight of Al in imure sample = (0.02074 g/0.8027 g)*100 = 2.60 %

queen_honey_blossom answered 2 hours ago

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