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Question

When 20.02 g of white solid (X) was heated, 4.4 g of acid gas (A) that turned lime water milky was driven off together with 1.8 g if a gas (B) which condenses to a colourless liquid. The solid (Y) that remained is dissolved in water to give an alkaline solution, with which excess barium chloride solution gave a white precipitate (Z). The precipitate effervesced with acid giving off carbon dioxide. Compound (A), (B) and (Y) are identified as:
(A) : CO2, (B) : H2O and (Y) : K2CO3
If true enter 1, else enter 0.

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Solution

The compound (X) is metal bicarbonate. On heating, it decomposes to form metal carbonate, carbon dioxide and water.
2MHCO32(m+61)gM2CO3+H2O+CO244g
Let M g/mol be tha molar mass of the metal. The molar mass of the metal carbonate will be M+61 g/mol. The molar mass of carbon dioxide is 44 g/mol.
2 moles of metal carbonate gives 1 mole of carbon dioxide.
2(M+61)20.02=444.4
Thus, M = 39.1.
Hence, the metal is potassium.
Thus, (X) is KHCO3, (Y) is K2CO3, (Z) is BaCO3, (A) is CO2 and (B) is H2O.
When potassium carbonate is dissolved in water to form alkaline solution, which is then treated with excess barium chloride solution a white ppt of barium carbonate is obtained which is compound (Z).
K2CO3+2H2O2KOH+H2O+CO2
CO2+BaCl2+H2OBaCO3+2HCl
Barium mcarbonate on treatment with acid forms carbon dioxide gas.
baCO3+2HClBaCl2+H2O+CO2

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