Aluminum hydroxide reacts with hydrochloric acid to produce water and aluminum chloride: Al(OH)3(aq) + 3HCl(aq) → 3H2O(l) + AlCl3(aq) Determine the grams of water formed by the reaction of 205 mL of 0.669 M aluminum hydroxide solution.
The Correct Answer and Explanation is:
To determine the grams of water produced from the reaction of 205 mL of 0.669 M aluminum hydroxide solution, follow these steps:
Step 1: Write the balanced chemical equation
Al(OH)3(aq)+3HCl(aq)→3H2O(l)+AlCl3(aq)\text{Al(OH)}_3(aq) + 3\text{HCl}(aq) \rightarrow 3\text{H}_2\text{O}(l) + \text{AlCl}_3(aq)Al(OH)3(aq)+3HCl(aq)→3H2O(l)+AlCl3(aq)
This shows that 1 mole of aluminum hydroxide forms 3 moles of water.
Step 2: Convert volume of solution to liters
205 mL=0.205 L205\ \text{mL} = 0.205\ \text{L}205 mL=0.205 L
Step 3: Use molarity to find moles of Al(OH)₃
Moles of Al(OH)3=0.669 mol/L×0.205 L=0.137145 mol\text{Moles of Al(OH)}_3 = 0.669\ \text{mol/L} \times 0.205\ \text{L} = 0.137145\ \text{mol}Moles of Al(OH)3=0.669 mol/L×0.205 L=0.137145 mol
Step 4: Use mole ratio to find moles of water formed
From the equation, 1 mol of Al(OH)₃ gives 3 mol of H₂O: Moles of H2O=0.137145 mol×3=0.411435 mol\text{Moles of H}_2\text{O} = 0.137145\ \text{mol} \times 3 = 0.411435\ \text{mol}Moles of H2O=0.137145 mol×3=0.411435 mol
Step 5: Convert moles of water to grams
Mass=moles×molar mass of H2O\text{Mass} = \text{moles} \times \text{molar mass of H}_2\text{O}Mass=moles×molar mass of H2O =0.411435 mol×18.015 g/mol=7.412 g= 0.411435\ \text{mol} \times 18.015\ \text{g/mol} = 7.412\ \text{g}=0.411435 mol×18.015 g/mol=7.412 g
Final Answer:
Approximately 7.41 grams of water are formed.
Explanation:
This problem involves stoichiometry based on a balanced chemical reaction. Aluminum hydroxide reacts with hydrochloric acid in a 1 to 3 mole ratio, producing aluminum chloride and water. The key is to first convert the volume of the solution into liters to work with molarity. Molarity gives moles per liter, so multiplying it by the volume in liters gives the actual number of moles of aluminum hydroxide available. Using the mole ratio from the balanced equation allows calculation of the moles of water produced. Finally, multiplying the moles of water by its molar mass (18.015 g/mol) gives the mass in grams. Each step applies basic chemistry principles of unit conversion and mole relationships.
