The ratio of the molar amounts of H2S needed to precipitate the metal ions form 20 ml each of 1 M Ca(NO3)2 and 0.5M CuSO4 is 1 : 1 2 : 1 1 : 2 indefinite
Question
The ratio of the molar amounts of H2S needed to precipitate the metal ions form 20 ml each of 1 M Ca(NO3)2 and 0.5M CuSO4 is 1 : 1 2 : 1 1 : 2 indefinite
Solution
The reaction of H2S with Ca(NO3)2 and CuSO4 can be represented as follows:
For Ca(NO3)2: Ca(NO3)2 + H2S -> CaS + 2HNO3
For CuSO4: CuSO4 + H2S -> CuS + H2SO4
From the balanced chemical equations, we can see that one mole of H2S reacts with one mole of Ca(NO3)2 or CuSO4 to form one mole of the respective metal sulfide.
Given that the molar concentration of Ca(NO3)2 is 1 M and the volume is 20 ml (or 0.02 L), the number of moles of Ca(NO3)2 is (1 mol/L) * (0.02 L) = 0.02 mol.
Similarly, given that the molar concentration of CuSO4 is 0.5 M and the volume is also 20 ml (or 0.02 L), the number of moles of CuSO4 is (0.5 mol/L) * (0.02 L) = 0.01 mol.
Therefore, the ratio of the molar amounts of H2S needed to precipitate the metal ions from the given solutions of Ca(NO3)2 and CuSO4 is 0.02 : 0.01, which simplifies to 2 : 1.
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