Analysis of a solid gave the following analysis results: (a) Treatment of the solid with dilute HCl gave off a gas, which when bubbled through \(\mathrm{a} \mathrm{Ca}(\mathrm{OH})_{2}\) solution gave a fine white precipitate. (b) A flame test gave a green flame. Deduce possible identities for the solid compound.

Short Answer

Expert verified
The solid compound can be Barium Carbonate, BaCO3.

Step by step solution

01

Analyze Clue from HCl Test

When HCl reacts with the solid to produce a white precipitate with \(\mathrm{Ca(OH)}_{2}\), it suggests that the gas liberated is carbon dioxide. Carbon dioxide forms a white precipitate of calcium carbonate when passed through a \(\mathrm{Ca(OH)}_{2}\) solution according to the reaction: \(\mathrm{Ca(OH)}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{CaCO}_{3} + \mathrm{H}_{2}\mathrm{O}\). This implies that the solid contains carbonate, CO3^2-.
02

Analyze Clue from Flame Test

The flame test producing a green flame indicates the presence of Barium, Ba2+. Barium compounds give a characteristic green color in the flame test.
03

Deduce the Identity of the Solid Compound

From the above analysis, it can be inferred that the solid is a compound containing Barium and Carbonate ions. Thus, the compound can be Barium Carbonate, BaCO3.

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