WAEC Chemistry 2019 Paper

Concept:The gas released from the reaction is identified by its characteristic reaction with lead(II)ethanoate paper.Explanation:Diluted hydrochloric acid reacts with iron(II)sulphide to produce iron(II)chloride and hydrogen sulphide gas.$$\mathrm{FeS} + 2\mathrm{HCl} \rightarrow \mathrm{FeCl}_2 + \mathrm{H}_2\mathrm{S}$$Hydrogen sulphide is the gas liberated.When this gas comes into contact with lead(II)ethanoate paper, lead(II)sulphide is formed.$$\mathrm{H}_2\mathrm{S} + (\mathrm{CH}_3\mathrm{COO})_2\mathrm{Pb} \rightarrow \mathrm{PbS} + 2\mathrm{CH}_3\mathrm{COOH}$$This lead(II)sulphide produces a black coloration on the paper.Therefore, the correct observation is that the gas turns lead(II)ethanoate paper black.The other options describe tests for different gases such as carbon dioxide, sulfur dioxide, or hydrogen.Answer:D. The gas turns lead(II)ethanoate paper black.