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A 0.2264 g sample of a pure carbonate, XnCO3(s) , was dissolved in 50.0 mL of 0.1800 M HCl(aq) . The excess HCl(aq) was back titrated with 24.90 mL of 0.0980 M NaOH(aq) . How many moles of HCl react with the carbonate

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Answer:

6.56x10⁻³ mol

Explanation:

First we calculate how many HCl moles are there in 50.0 mL of a 0.1800 M solution:

  • 0.1800 M * 50.0 mL = 9.00 mmol HCl

Then we need to calculate how many HCl moles were in excess, that number is the same as the number of NaOH moles they reacted with:

  • 0.0980 M * 24.90 mL = 2.44 mmol NaOH = 2.44 mmol HCl

Finally we calculate the difference between the original number of HCl moles and the number remaining after the reaction with XnCO₃:

  • 9.00 mmol - 2.44 mmol = 6.56 mmol
  • 6.56 mmol / 1000 = 6.56x10⁻³ mol

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