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If 1.6 gm of so2 and 1.5 ×10^22 molevules of h2s are mixed and allowed to remain in contact in aclosed vessel until tue reaction proceeds to complete . Whuch is true?
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If 1.6 gm of so2 and 1.5 ×10^22 molevules of h2s are mixed and allowed...
Given information:

  • Amount of SO2 = 1.6 gm

  • Number of molecules of H2S = 1.5 × 10^22



Reaction to be considered:

  • SO2 + 2H2S → 3S + 2H2O



Determining the limiting reactant:

  • Number of moles of SO2 = 1.6 g / 64 g/mol = 0.025 mol

  • Number of moles of H2S = 1.5 × 10^22 / Avogadro's Number = 2.5 × 10^-2 mol

  • From the balanced chemical equation, it is clear that 1 mole of SO2 reacts with 2 moles of H2S.

  • Therefore, the amount of H2S required to react with 0.025 mol of SO2 would be 0.025 × 2 = 0.05 mol.


Since the amount of H2S available (0.025 mol) is less than the required amount (0.05 mol), H2S is the limiting reactant.


Calculating the amount of product formed:

  • From the balanced chemical equation, it is clear that 1 mole of SO2 reacts with 3 moles of S.

  • Therefore, the amount of S formed would be 3 × 0.025 = 0.075 mol.

  • Mass of S formed = 0.075 × 32 g/mol = 2.4 g.



Conclusion:
Thus, it can be concluded that 2.4 g of S will be formed when 1.6 gm of SO2 and 1.5 × 10^22 molecules of H2S are mixed and allowed to react completely.
Community Answer
If 1.6 gm of so2 and 1.5 ×10^22 molevules of h2s are mixed and allowed...
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If 1.6 gm of so2 and 1.5 ×10^22 molevules of h2s are mixed and allowed to remain in contact in aclosed vessel until tue reaction proceeds to complete . Whuch is true?
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