Answer:
1.596 g
Step-by-step explanation:
The neutralization reaction is:
NaHCO₃ + HCl → NaCl + H₂O + CO₂
The number of moles of the acid in stomach is the volume (250.0 mL = 0.250 L) multiplied by the molar concentration (0.076 M):
n = 0.250 * 0.076 = 0.019 mol of HCl
By the stoichiometry of the reaction, 1 mol of NaHCO₃ is needed to neutralize 1 mol of HCl, so it'll be necessary 0.019 mol of sodium bicarbonate.
The molar mass of NaHCO₃ is 84 g/mol, thus the mass of it is:
m = number of moles * molar mass
m = 0.019 * 84
m = 1.596 g