Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily \(\mathrm{HCl}\) : \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (a) Balance this equation. (b) Calculate the number of grams of \(\mathrm{HCl}\) that can react with \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) (c) Calculate the number of grams of \(\mathrm{AlCl}_{3}\) and the number of grams of \(\mathrm{H}_{2} \mathrm{O}\) formed when \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

For the given reaction, the chemical equation is: Al(OH)₃(s) + HCl(aq) → AlCl₃(aq) + H₂O(l) To balance this equation, we can follow these steps: 1. Count the number of each atom in the reactants and products side. 2. Adjust the coefficients (the numbers in front of the compounds) to have an equal number of atoms on both sides. The balanced chemical equation is: Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3H₂O(l) Now we proceed to find the mass of HCl and products.

We're given that the mass of Al(OH)₃ is 0.500 g. We need to first find the moles of Al(OH)₃. To determine the moles, we'll use the formula: Moles of Al(OH)₃ = (Mass of Al(OH)₃) / (Molar mass of Al(OH)₃) The molar mass of Al(OH)₃ is (27.0 g/mol for Al) + (3 x 17.0 g/mol for OH) = 78.0 g/mol. Moles of Al(OH)₃ = 0.500 g / 78.0 g/mol = 0.00641 mol Using stoichiometry and the balanced chemical equation, we can now find the moles of HCl required to react with 0.500 g of Al(OH)₃: 1Al(OH)₃ reacts with 3HCl, so for 0.00641 mol of Al(OH)₃, we need: (0.00641 mol Al(OH)₃) * (3 mol of HCl / 1 mol Al(OH)₃) = 0.0192 mol of HCl Finally, we find the mass of this amount of HCl using its molar mass (1.0 g/mol for H and 35.5 g/mol for Cl): Mass of HCl = (0.0192 mol) * (36.5 g/mol) = 0.701 g of HCl

We already determined that 0.00641 mol of Al(OH)₃ reacted. We can use stoichiometry to find the moles of AlCl₃ and H₂O formed. For AlCl₃, the ratio is 1:1 for Al(OH)₃:AlCl₃. Moles of AlCl₃ = 0.00641 mol For H₂O, the ratio is 1:3 for Al(OH)₃:H₂O. Moles of H₂O = 0.00641 mol * (3 mol H₂O / 1 mol Al(OH)₃) = 0.0192 mol of H₂O Now, we can find the mass of AlCl₃ and H₂O using their respective molar masses. Mass of AlCl₃ = (0.00641 mol) * (133.5 g/mol) = 0.856 g of AlCl₃ Mass of H₂O = (0.0192 mol) * (18.0 g/mol) = 0.346 g of H₂O

The law of conservation of mass states that the mass of reactants should be equal to the mass of products. In this case, the mass of Al(OH)₃ and HCl should be equal to the mass of AlCl₃ and H₂O. Reactants mass = Mass of Al(OH)₃ + Mass of HCl = 0.500 g + 0.701 g = 1.201 g Products mass = Mass of AlCl₃ + Mass of H₂O = 0.856 g + 0.346 g = 1.202 g There might be a small discrepancy between the mass of reactants and products due to rounding, but it is within an acceptable range. Thus, our calculations are consistent with the law of conservation of mass, and the solution is complete.