Final answer:
The mass of the precipitate Ni(OH)₂ formed when 20.5 mL of 0.300 M Ni(NO₃)₂ reacts with 35.0 mL of 0.300 M NaOH is 0.570 grams.
Step-by-step explanation:
To determine the mass of precipitate formed when 0.300 M Ni(NO₃)₂ reacts with 0.300 M NaOH, we first need to use stoichiometry based on the balanced chemical equation:
Ni(NO₃)₂(aq) + 2NaOH(aq) → Ni(OH)₂(s) + 2NaNO₃(aq)
Firstly, we need to calculate the number of moles of Ni(NO₃)₂ and NaOH that are reacting. Using the molarity and volume of each solution:
- Moles of Ni(NO₃)₂ = Molarity of Ni(NO₃)₂ × Volume of Ni(NO₃)₂ = 0.300 mol/L × 0.0205 L = 0.00615 mol
- Moles of NaOH = Molarity of NaOH × Volume of NaOH = 0.300 mol/L × 0.0350 L = 0.0105 mol
Since the reaction requires two moles of NaOH for every mole of Ni(NO₃)₂, the limiting reagent is Ni(NO₃)₂ (because it will be completely consumed first).
The number of moles of Ni(OH)₂ precipitate formed will be equal to the number of moles of Ni(NO₃)₂, which is 0.00615 mol. The molar mass of Ni(OH)₂ is approximately 92.71 g/mol, so we can calculate the mass:
Mass of Ni(OH)₂ = Moles of Ni(OH)₂ × Molar mass of Ni(OH)₂ = 0.00615 mol × 92.71 g/mol = 0.570 g
Hence, 0.570 grams of Ni(OH)₂ precipitate will form.