Anonymous
Anonymous asked in Science & MathematicsChemistry · 4 weeks ago

A 6.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0?

A 6.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0 molar HCL solution 

A 249 milliliter sample of dry CO2 gas, measured at 22*C and 740 mmHg, is obtained from the reaction. What is the percentage of potassium carbonate in the mixture?

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  • 4 weeks ago
    Best Answer

    Supposing the given amount of HCl is sufficient to react with all of the KOH and K2CO3:

    n = PV/RT = (740 mmHg) x (0.249 L) / ((62.36367 L mmHg/K mol) x (22 + 273) K) = 0.01002 mol CO2

    K2CO3 + 2 HCl → 2 KCl + CO2 + H2O

    (0.01002 mol CO2) x (1 mol K2CO3 / 1 mol CO2) x (138.2055 g K2CO3/mol) / (6.00 g) = 0.230803 =

    23.1% K2CO3

  • Anonymous
    4 weeks ago

    Its very simple. I will give you the answer, but you first have to choose my answer as best answer then I will edit my answer and put the answer to your question showing full calculations as well.

    • Bobby_Thin
      Lv 7
      4 weeks agoReport

      Nice idea but once BA is awarded the question is closed and cannot be edited. .

      There should be some way of rewarding the answerer ... the asker is under no pressure to respond

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