Atomic StructureHard
Question
Which of the following statements is/are correct ?
1.0 g mixture of CaCO3(s) and glass beads liberate 0.22 g of CO2 upon treatment with excess of HCl. Glass does not react with HCl.
CaCO3 + 2HCl → CO2 + H2O + CaCl2
[Mw CaCO3 = 100, Mw of CO2 = 44, [Atomic weight of Ca = 40]
1.0 g mixture of CaCO3(s) and glass beads liberate 0.22 g of CO2 upon treatment with excess of HCl. Glass does not react with HCl.
CaCO3 + 2HCl → CO2 + H2O + CaCl2
[Mw CaCO3 = 100, Mw of CO2 = 44, [Atomic weight of Ca = 40]
Options
A.The weight of CaCO3 in the original mixture is 0.5 g.
B.The weight of calcium in the original mixture is 0.2 g,
C.The weight percent of calcium in the original mixture is 40% Ca.
D.The weight percent of Ca in the original mixture is 20% Ca.
Solution
a. Weight of CaCO3 = (0.22g CO2)
= 0.5 g CaCO3
b. Moles of CaCO3 = moles of Ca
=
= 0.005 mol
weight Ca = 0.005 × 40 = 0.2 g Ca
d. % of Ca =
× 100 = 20% Ca
Hence, (c) is wrong.
= 0.5 g CaCO3b. Moles of CaCO3 = moles of Ca
=
= 0.005 molweight Ca = 0.005 × 40 = 0.2 g Ca
d. % of Ca =
× 100 = 20% CaHence, (c) is wrong.
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