2 gram mixture of NaHCO3 and KCLO3 requires 100ml of 0.1N HCL for complete neutralization.The weight of KCLO3 present in the mixture is
Dear student,
As the mixture contains KClO3 and NaHCO3, Out of these two KClO3 is neutral as it is a salt of strong acid and strong base while NaHCO3 being salt of strong base and weak acid forms basic solution.
As NaHCO3 is basic, it will participate in complete neutralization with acid.
Given, amount of mixture = 2 g
Normality (or Molarity) of HCl = 0.1 N
Volume of HCl = 100 mL or 0.1 L
Moles of HCl = molarity x volume = 0.1 x 0.1 = 0.01 moles
Let amount of NaHCO3 = x g
Amount of KClO3 = 2-x g
Molar mass of NaHCO3 = 84 g
Moles of NaHCO3 = x/84
As NaHCO3 reacts with HCl in equimolar amount i.e. 1 mole of NaHCO3 reacts with 1 mole of HCl, then
Moles of NaHCO3 must equal moles of HCl
So, x/84 = 0.01
x = 0.84 g
Amount of KClO3 = 2-0.84 = 1.16 g
Regards.
As the mixture contains KClO3 and NaHCO3, Out of these two KClO3 is neutral as it is a salt of strong acid and strong base while NaHCO3 being salt of strong base and weak acid forms basic solution.
As NaHCO3 is basic, it will participate in complete neutralization with acid.
Given, amount of mixture = 2 g
Normality (or Molarity) of HCl = 0.1 N
Volume of HCl = 100 mL or 0.1 L
Moles of HCl = molarity x volume = 0.1 x 0.1 = 0.01 moles
Let amount of NaHCO3 = x g
Amount of KClO3 = 2-x g
Molar mass of NaHCO3 = 84 g
Moles of NaHCO3 = x/84
As NaHCO3 reacts with HCl in equimolar amount i.e. 1 mole of NaHCO3 reacts with 1 mole of HCl, then
Moles of NaHCO3 must equal moles of HCl
So, x/84 = 0.01
x = 0.84 g
Amount of KClO3 = 2-0.84 = 1.16 g
Regards.