Question

The amount of $ {{\text{H}}_{\text{2}}}\text{S}\text{.} $ required to precipitate $ \text{1}\text{.69 g BaS} $ from $ \text{BaC}{{\text{l}}_{\text{2}}} $ solution is:

• A. 3.4 g
• B. 0.034 g
• C. 0.34 g
• D. 0.17g

Answer 1

The Correct Option is C

Key Idea: First write balanced chemical reaction and then find the answer. $ BaC{{l}_{2}}+{{H}_{2}}S\xrightarrow{{}}BaS+2HCl $ Molecular mass of $ {{H}_{2}}S=1\times 2+32=34 $ Molecular mass of $ BaS=137+32=169 $ Given mass of $ BaS=1.69\,g $ According to reaction. $ \because $ $ 169\,g $ of $ \text{BaS} $ is obtained by $ \text{=}\,\text{34}\,\text{g}\, $ of $ {{H}_{2}}S $ $ \therefore $ 1.69 g of $ \text{BaS} $ is obtained by $ =\frac{34}{169}\times 1.69 $ $ =0.34\,g\,\text{of}\,{{\text{H}}_{\text{2}}}\text{S} $