Answer:
The mass of aluminum hydroxide that will precipitate when 35.5 mL of 0.145 mol/L aqueous aluminum nitrate is mixed with 44.2 mL of 0.215 mol/L aqueous sodium hydroxide is approximately 0.247 grams
Explanation:
The given parameters are;
The volume of 0.145 mol/L aqueous aluminum nitrate in the reaction = 35.5 mL
The volume of 0.215 mol/L aqueous sodium hydroxide in the reaction = 44.2 mL
The balanced chemical equation for the reaction is given as follows;
Al(NO₃)₃ (aq) + 3NaOH (aq) → Al(OH)₃ (s) + 3NaNO₃ (aq)
Therefore;
1 mole of Al(NO₃)₃, reacts with 3 moles of NaOH to form 1 mole Al(OH)₃, and 3 moles of NaNO₃
The number of moles of aqueous aluminum nitrate in the reaction = 35.5/1000 L × 0.145 mol/L = 0.0051475 moles
The number of moles of aqueous sodium hydroxide in the reaction = 44.2/1000 L × 0.215 mol/L = 0.009503 moles
Therefore, 0.009503 moles NaOH will react with 0.009503/3 moles Al(NO₃)₃ to form 0.009503/3 moles of aluminum hydroxide precipitate
The molar mass of aluminum hydroxide = 78 g/mol
Therefore, 0.009503/3 moles of aluminum hydroxide weighs 78 × 0.009503/3 = 0.247078 grams
The mass of aluminum hydroxide that will precipitate out ≈ 0.247 grams.
