References A 4.470 g sample of a mixture of sodium hydrogen carbonate and potass

Question

References A 4.470 g sample of a mixture of sodium hydrogen carbonate and potassium chloride is dissolved in 25.50 mL of 0.444 M H SO4 Some acid remains after treatment of the sample. If 35.2 ml of were required to titrate the excess sulfuric acid, how many moles of sodium hydrogen carbonate were present in the onginal sample? 0.103 M NaOH mol NaHCO What is the percent composition of the original sample? o NaHCO KCl Submit Answer Try Another Version 3 item attempts remaining Previcus Next Emall Instructor Save and

Explanation / Answer

Moles of H2SO4 = 25.5 x 0.444 /1000 = 0.011322 Moles

Moles of NaOH reacted against H2SO4 = 35.2 x 0.103 / 1000 =  0.0036256 Moles

Moles of H2SO4 reacted = 0.011322 - 0.0018128 =  0.0095092 Moles

Moles of NaHCO3 were present in the original sample = 0.0095092 Moles

Mass of NaHCO3 =  0.0095092 x 84 = 0.79877 gm

Mass of KCl = 4.47 - 0.79877 = 3.6712 gm

Mass % KCl = 3.8235 x 100 / 4.47 = 82.13 %

Mass % NaHCO3 = 100 -85.53 = 17.87 %

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