A student adds an excess of 32. 36mL of 0.1034 M HCL(aq) to 0.326h of a powdered

Question

A student adds an excess of 32. 36mL of 0.1034 M HCL(aq) to 0.326h of a powdered anti Acid. The student boils the mixture and then allows it to cool. A. The student then titrates the mixture to the end points using 11.72ml 0.1506 M NaOH(aq). Calculate number of moles of HCL neutralized by the NaOH added. B.calculate the number of moles of HCL that were neutralized by the antacid. Note that this quantity equals the number of equivalents of antacid A student adds an excess of 32. 36mL of 0.1034 M HCL(aq) to 0.326h of a powdered anti Acid. The student boils the mixture and then allows it to cool. A. The student then titrates the mixture to the end points using 11.72ml 0.1506 M NaOH(aq). Calculate number of moles of HCL neutralized by the NaOH added. B.calculate the number of moles of HCL that were neutralized by the antacid. Note that this quantity equals the number of equivalents of antacid A. The student then titrates the mixture to the end points using 11.72ml 0.1506 M NaOH(aq). Calculate number of moles of HCL neutralized by the NaOH added. B.calculate the number of moles of HCL that were neutralized by the antacid. Note that this quantity equals the number of equivalents of antacid

Explanation / Answer

A. number of moles of HCL neutralized by the NaOH added = M*V/1000

   = 11.72*0.1506/1000

   = 1.76*10^-3 mol

B. number of moles of HCL that were neutralized by the antacid

      = (32.36*0.1034/1000)-(1.76*10^-3)

      = 1.58*10^-3 mol

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