How many mg of NaHCO3 are in a 500-mg tablet if 40.00 mL of 0.120 M HCl is required to neutralize the sample?
Answer:
Moles HCl = 40.00 x 0.120 /1000 = 0.0048 = moles NaHCO3
Molar mass NaHCO3 = 84 g/mol
0.0048 mol x 84 g/mol = 0.403 g = 403 mg
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