Asked by Science
The approximate concentration of hydrochloric acid HCL, in the stomach (stomach acid) is 0.17M, calculate the mass of the following antacid required to neutralize 50cm of this acid; bicarbonate of soda, NaHCO3
Answers
Answered by
DrBob222
I'm sure you meant 50 cc or 50 cm^3 but NOT 50 cm.
NaHCO3 + HCl ==> NaCl + H2O + CO2
millimoles HCl = mL x M = 50 mL x 0.17 = approx 85
millimoles NaHCO3 neutralized = approx 85 = 0.085 moles
grams NaHCO3 = moles NaHCO3 x molar mass NaHCO3 = ? grams.
NaHCO3 + HCl ==> NaCl + H2O + CO2
millimoles HCl = mL x M = 50 mL x 0.17 = approx 85
millimoles NaHCO3 neutralized = approx 85 = 0.085 moles
grams NaHCO3 = moles NaHCO3 x molar mass NaHCO3 = ? grams.
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