Aluminum nitrate reacts with sodium hydroxide to form a precipitate of aluminum hydroxide. What mass of aluminum hydroxide 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?

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.

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