A solution of 300.0 mL of 0.150 M Nickel (II) sulfate is mixed with a solution of 100.0 mL of 0.200 M potassium hydroxide.

Question

A solution of 300.0 mL of 0.150 M Nickel (II) sulfate is mixed with a solution of 100.0 mL of 0.200 M potassium hydroxide.

(a) Write the balanced chemical equation for the reaction that occurs.

(b) Identify the limiting reactant. 

(c) How many grams of the precipitate form?

(d) What is the concentration of precipitate in the solution? 

Answer 1

(a) Balanced equation: NiSO_{4 (aq)} + 2KOH_(aq) → Ni(OH)_{2 (s)} + K₂SO_{4 (aq)}

(b) Find the limiting reactant:

moles NiSO₄ = (300.0mL)(1L/1000mL)(0.150moles/L) = 0.045 moles NiSO₄

moles KOH = (100.0mL)(1L/1000mL)(0.200moles/L) = 0.020 moles KOH ← limiting reactant

*the balanced equation shows us that 1 mole of NiSO₄ reacts with 2 moles of KOH, so 0.045 moles NiSO₄ need 0.090 moles of KOH to fully react. Since there are only 0.020 moles of KOH in the reaction, KOH limits the reaction.

(c) Ni(OH)₂ precipitates out of the solution. We use the limiting reactant to find the theoretical mass of the precipitate. Looking at the equation, 2 moles of KOH reacts to produce 1 mole of Ni(OH)₂.

Therefore, (0.020 moles KOH)(1 mole Ni(OH)₂ / 2 moles KOH) = 0.010 moles Ni(OH)₂

Convert moles Ni(OH)₂ to grams using the molar mass:

(0.010 moles)(92.708 grams/mole Ni(OH)₂) = 0.927g Ni(OH)₂

(d) What is the concentration of Ni(OH)₂ in the solution?

We know that there is 0.010 moles Ni(OH)₂ and we use the total volume of the solution to find the molarity.

300.0mL + 100.0mL = 400.0mL then we divide moles by liters (400.0mL=0.400L)

(0.010 moles Ni(OH)₂) / (0.400L) = 0.025 M Ni(OH)₂