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Chemistry Question on Expressing Concentration of Solutions

The amount of H2S.{{\text{H}}_{\text{2}}}\text{S}\text{.} required to precipitate 1.69 g BaS\text{1}\text{.69 g BaS} from BaCl2\text{BaC}{{\text{l}}_{\text{2}}} solution is:

A

3.4 g

B

0.034 g

C

0.34 g

D

0.17g

Answer

0.34 g

Explanation

Solution

Key Idea: First write balanced chemical reaction and then find the answer. BaCl2+H2S→BaS+2HClBaC{{l}_{2}}+{{H}_{2}}S\xrightarrow{{}}BaS+2HCl Molecular mass of H2S=1×2+32=34{{H}_{2}}S=1\times 2+32=34 Molecular mass of BaS=137+32=169BaS=137+32=169 Given mass of BaS=1.69 gBaS=1.69\,g According to reaction. ∵\because 169 g169\,g of BaS\text{BaS} is obtained by = 34 g \text{=}\,\text{34}\,\text{g}\, of H2S{{H}_{2}}S ∴\therefore 1.69 g of BaS\text{BaS} is obtained by =34169×1.69=\frac{34}{169}\times 1.69 =0.34 g of H2S=0.34\,g\,\text{of}\,{{\text{H}}_{\text{2}}}\text{S}