Final answer:
Approximately 0.29168 grams of hydrochloric acid (HCl) can be neutralized by 0.400 g of calcium carbonate (CaCO3) based on the stoichiometry of the neutralization reaction where one mole of CaCO3 can neutralize two moles of HCl.
Step-by-step explanation:
To calculate the number of grams of acid that can be neutralized by 0.400 g of CaCO3, we must first understand the neutralization reaction that takes place. Calcium carbonate (CaCO3) reacts with stomach acid, which is primarily hydrochloric acid (HCl), following the balanced chemical equation:
CaCO3 + 2 HCl → CaCl2 + CO2 + H2O
From the equation, one mole of CaCO3 neutralizes two moles of HCl. The molar mass of CaCO3 is roughly 100.09 g/mol. If we have 0.400 g of CaCO3, we can calculate the moles of CaCO3:
moles of CaCO3 = grams of CaCO3 / molar mass = 0.400 g / 100.09 g/mol = 0.004 mol
Next, since 1 mole of CaCO3 neutralizes 2 moles of HCl, we determine the moles of HCl that can be neutralized:
moles of HCl neutralized = 2 × moles of CaCO3 = 2 × 0.004 mol = 0.008 mol
To find the gram equivalent of HCl, we need its molar mass, which is about 36.46 g/mol:
grams of HCl = moles × molar mass = 0.008 mol × 36.46 g/mol = 0.29168 g
Therefore, 0.400 g of CaCO3 could neutralize approximately 0.29168 g of hydrochloric acid (HCl).