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%
"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.
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