Calcium carbonate reacts with stomach acid (HCI) according to the following equation: Caco, (s) + 2 HCl(aq) - CaCl, (aq) + CO2(g) + H20 (1)
Tums, an antacid, contains CaCo3. If Tums is added to 25.7 mL of a 0.738 M HCl solution, how many grams of Co, are produced?

0.418 g of CO₂ will be produced

Explanation:

Equation of the reaction is given below:

CaCO₃ (s) + 2HCl (aq) > CaCl₂(aq) + CO₂ (g) + H₂O (l)

From the equation of reaction , 1 mole of CaCO₃ reacts with 2 molemolesmof HCl to produce 1 mole of CO₂

Number of moles of HCl in 25.7 ml of a 0.738 M solution of HCl is obtained using the formula below:

Number of moles = molarity × volume (Litres)

Number of moles of HCl = 0.738 M × 25.7 mL × 1 L/1000 mL = 0.0190 moles

Since the antacid, Tums will be excess, the limiting reactant is HCl

2 moles of HCl produces 1 mole of CO₂

0.0190 moles of HCl will produce 0.0190 moles ÷ 2 = 0.0095 molesof CO₂

Mass of 0.0095 moles of CO₂ = Numbe of moles × molar mass of CO₂

Molar mass of CO₂ = 44.0 g/mol

Mass of CO₂ produced = 0.00950 moles × 44.0 g/mol = 0.418 g of CO₂

Therefore, 0.418 g of CO₂ will be produced if Tums is added to 25.7 mL of a 0.738 M HCl solution

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.