Analysis of antacids Total NaOH 39.45 mL, Total Hcl 99.70mL , mole of HCl were a

Question

Analysis of antacids

Total NaOH 39.45 mL,Total Hcl 99.70mL , mole of HCl were added 0.000982 moles ,moles of NaOH were added 0.00404moles,Hcl React with NaOH 0.00404,Hcl React with sodium carbonate 0.00578mole, Sodium Carbonat React with HCl2.89*10^-5 mole,mole of Carbonate =1 mole,Mass of CO3^-2 =1.74g,the % bymass of carbonate 22.8%

Class Cost Analysis for Commercial Antacids Analysis of antacids Total NaOH 39.45 mL,Total Hcl 99.70mL , mole of HCl were added 0.000982 moles ,moles of NaOH were added 0.00404moles,Hcl React with NaOH 0.00404,Hcl React with sodium carbonate 0.00578mole, Sodium Carbonat React with HCl2.89*10^-5 mole,mole of Carbonate =1 mole,Mass of CO3^-2 =1.74g,the % bymass of carbonate 22.8% Table 2.

Explanation / Answer

mass of tablet=xg

Initial Moles of HCl added=molarity(moles/L)*Volume(L)

moles of NaOH used to titrate excess HCl=Molarity of NaOH(moles/L)*Volume(L)

Moles of HCl used against Carbonate=total moles of HCl added-moles of HCl titrated against NaOH

Moles of Carbonate=0.5*moles of HCl used against carbonate

mass of carbonate=moles*molar mass

%age mass of carbonate=mass determined/total mass of tablet*100

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