A 3.850 g sample of a mixture of sodium hydrogen carbonate and potassium chlorid

Question

A 3.850 g sample of a mixture of sodium hydrogen carbonate and potassium chloride is dissolved in 25.10 mL of 0.445 M H2SO4. Some acid remains after treatment of the sample. Write both the net ionic and the molecular equations for the complete reaction of sodium hydrogen carbonate with sulfuric acid. If 33.9 mL of 0.105 M NaOH were required to titrate the excess sulfuric acid, how many moles of sodium hydrogen carbonate were present in the original sample? What is the percent composition of the original sample?

- mol NaHCO3

- % KCl

Please show work

Explanation / Answer

   Molecular equation : 2NaHCO3(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l) + 2co2(g)

Net ionic equation :   HCO3^-(aq) + H+(aq)   ----> H2O(l) + co2(g)

From equation:

2 mole NaHCO3 = 1 mole H2SO4 = 1 mole Na2SO4 =

No of moles of H2SO4 added = (25.1/1000)*0.445 = 0.01117 mole

No of moles of H2SO4 excess = 1/2* No of moles of NaOH

                           = (1/2)*(33.9/1000)*0.105

                        = 0.00178 mole

No of moles of H2SO4 reacted = 0.01117 - 0.00178

   = 0.00939 mole

No of mole NaHCO3 present in the original sample

      = 0.00939*2 = 0.01878 mole

molarmass of NaHCO3 = 84.007 g/mol

mass of NaHCO3 = no of moles * Mwt = 0.01878*84.007 = 1.58 grams

percent composition of NaHCO3 = w/total mass*100

   = 1.58/3.85*100 = 41.039%

percent composition of KCl = (3.85-1.58)/(3.85)*100   = 58.9%

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