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

Answer 1

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.