Question
"The percentage of sodium hydrogen carbonate, NaHCO3, in a powder for stomach upsets is found by titrating .275 M hydrochloric acid. if 15.5 mL of hydrochloric acid is required to react with .500 g of the sample, what is the percentage of sodium hydrogen carbonate in the sample?
the balanced equation for the reaction that takes place is
NaHCO3 (s) +H^+(aq) -> Na^+(aq) + CO2(g) + H2O "
why is H^+ used in the rxn and not HCl (aq)? And how do i start the prob.?
HCL and NaCl are (aq) spectator ions, not needed. If you feel a need, put them in, it wont change the reaction.
How many moles of H+ are used? That is the same number of moles of sodium bicarbonate. Compute the grams of sodium bicarbonate that equates to, then you can do the percent.
the balanced equation for the reaction that takes place is
NaHCO3 (s) +H^+(aq) -> Na^+(aq) + CO2(g) + H2O "
why is H^+ used in the rxn and not HCl (aq)? And how do i start the prob.?
HCL and NaCl are (aq) spectator ions, not needed. If you feel a need, put them in, it wont change the reaction.
How many moles of H+ are used? That is the same number of moles of sodium bicarbonate. Compute the grams of sodium bicarbonate that equates to, then you can do the percent.
Answers
15.5 mL of .275 M HCl = 15.5 mL x .275 mol/L = 4.2875 moles H+
4.2875 moles NaHCO3 = 4.2875 moles x 84.007 g/mol = 358.9 g NaHCO3
Percentage of NaHCO3 = (358.9 g/500 g) x 100 = 71.8%
4.2875 moles NaHCO3 = 4.2875 moles x 84.007 g/mol = 358.9 g NaHCO3
Percentage of NaHCO3 = (358.9 g/500 g) x 100 = 71.8%
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