Final answer:
The mass of Ni(OH)₂ precipitate formed when 20.5 mL of 0.300 M Ni(NO₃)₂ reacts with 32.5 mL of 0.300 M NaOH is 0.5700 grams.
Step-by-step explanation:
To determine the mass of the precipitate formed when 0.300 M Ni(NO₃)₂ reacts with 0.300 M NaOH, the balanced equation Ni(NO₃)₂(aq) + 2 NaOH(aq) → Ni(OH)₂(s) + 2 NaNO₃(aq) is used. The reaction shows that 1 mole of Ni(NO₃)₂ reacts with 2 moles of NaOH to form 1 mole of Ni(OH)₂ precipitate.
To calculate the number of moles of Ni(NO₃)₂ used, multiply the volume in liters (0.0205 L) by the molarity (0.300 M). Similarly, calculate the moles of NaOH (0.0325 L * 0.300 M). Since NaOH is in excess (0.00975 moles compared to 0.00615 moles Ni(NO₃)₂), the limiting reactant is Ni(NO₃)₂.
Number of moles of Ni(OH)₂ formed is equal to the moles of Ni(NO₃)₂ (limiting reactant), which is 0.00615 moles. The molar mass of Ni(OH)₂ is approximately 92.71 g/mol. Therefore, multiplying the number of moles of Ni(OH)₂ by its molar mass gives the mass of Ni(OH)₂ precipitate: 0.00615 moles * 92.71 g/mol = 0.5700 grams.