Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily 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 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.

The unbalanced equation given is: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ To balance the equation, we need to have the same number of atoms for each element on both sides. The balanced equation is: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+3\mathrm{H}_{2} \mathrm{O}(l) $$

First, we need to calculate the moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) given that its mass is 0.500 grams. We use the formula: moles = (mass)/(molar mass). The molar mass of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(26.98 + 3(15.999) + 3(1.008) = 78.00\) g/mol. So, the moles are: $$ \mathrm{moles} = \frac{0.500\ \text{g}}{78.00\ \text{g/mol}} = 6.41 \times 10^{-3}\ \text{mol} $$ Now, from the balanced equation, we see that one mole of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts with three moles of HCl. Therefore, the moles of HCl required are: $$ \mathrm{moles\:of\:HCl} = 3 \times 6.41 \times 10^{-3}\ \text{mol} = 1.92 \times 10^{-2}\ \text{mol} $$ Finally, we find the mass of HCl using the formula: mass = (moles) × (molar mass). The molar mass of HCl is \(1.008 + 35.45 = 36.46\) g/mol. Thus, the mass of HCl is: $$ \mathrm{mass\:of\:HCl} = (1.92 \times 10^{-2}\ \text{mol})(36.46\ \text{g/mol}) = 0.700 \ \text{g} $$

The balanced equation shows that 1 mole of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts to produce 1 mole of \(\mathrm{AlCl}_{3}\) and 3 moles of \(\mathrm{H}_{2}\mathrm{O}\). Using the moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) calculated previously, we can find the moles of the products: Moles of \(\mathrm{AlCl}_3\): \(6.41 \times 10^{-3}\ \text{mol}\) Moles of \(\mathrm{H}_{2}\mathrm{O}\): \(3 \times 6.41 \times 10^{-3}\ \text{mol} = 1.92 \times 10^{-2}\ \text{mol}\) Now, let's calculate the mass of the products: Molar mass of \(\mathrm{AlCl}_3\): \(26.98 + 3(35.45) = 133.34\) g/mol Molar mass of \(\mathrm{H}_{2}\mathrm{O}\): \(2(1.008) + 15.999 = 18.015\) g/mol Mass of \(\mathrm{AlCl}_3\): \((6.41 \times 10^{-3}\ \text{mol})(133.34\ \text{g/mol}) = 0.854\ \text{g}\) Mass of \(\mathrm{H}_{2}\mathrm{O}\): \((1.92 \times 10^{-2}\ \text{mol})(18.015\ \text{g/mol}) = 0.346\ \text{g}\)

According to the law of conservation of mass, the total mass of reactants must equal the total mass of products. Let's check if our calculations conform to this rule: Reactants: Mass of \(\mathrm{Al}(\mathrm{OH})_{3}\) + Mass of HCl \(0.500\ \text{g} + 0.700\ \text{g} = 1.200\ \text{g}\) Products: Mass of \(\mathrm{AlCl}_{3}\) + Mass of \(\mathrm{H}_{2}\mathrm{O}\) \(0.854\ \text{g} + 0.346\ \text{g} = 1.200\ \text{g}\) Since the mass of reactants equals the mass of products, our calculations are consistent with the law of conservation of mass.