5.0g sample of calcium carbonate (caco³) contaminated with a volitile impurity ,left a residue of 2.2g on strong heating .what is the percentage of pure caco³ in the sample ? [ Ca =40,c=12,o=16]
Answers
Answered by
1
Answer:
CaCO
3
⇌
Δ
CaO+CO
2
Let pure sample of CaCO
3
=x grams
Mass of CaO produced after decomposition =22.4 g
Molar mass of CaCO
3
=100 g/mol
and, Molar mass of CaO=56 g
If 100% is pure, then
100 g CaCO
3
⟶56 g of CaO
Also,y g of CaCO
3
⟶22.4 g of CaO
Dividing these two,
y
100
=
22.4
56
y=40 grams
∴ Percentage of purity =
Totalmassofimpure
Massofpuresample
×100=
50
40
×100=80%
Explanation:
Similar questions