A solution of 220mL of 0.200M Koh is mixed with a solution of 120mL of 0.240M NiSO4.

1.Determine limiting reactant?
2.Calculate the mass of precipitate formed?
3.Calculate the molarity of all ions left in the solution?

See your other post above.

To determine the limiting reactant, we need to compare the amount of each reactant relative to their stoichiometric ratios in the balanced equation.

1. First, let's write the balanced equation for the reaction between KOH and NiSO4:
KOH + NiSO4 --> K2SO4 + Ni(OH)2

In the equation, the ratio between KOH and NiSO4 is 1:1. This means that one mole of KOH reacts with one mole of NiSO4.

2. Now, let's find the number of moles of each reactant:
The volume of the KOH solution is given as 220 mL (0.220 L), and the molarity is given as 0.200 M. Therefore, the number of moles of KOH can be calculated as follows:
moles of KOH = volume (in L) x molarity = 0.220 L x 0.200 M = 0.044 moles

Similarly, the number of moles of NiSO4 can be calculated using the given volume (120 mL or 0.120 L) and molarity (0.240 M):
moles of NiSO4 = volume (in L) x molarity = 0.120 L x 0.240 M = 0.0288 moles

3. Now we compare the moles of each reactant to identify the limiting reactant:
From the balanced equation, we know that the stoichiometric ratio between KOH and NiSO4 is 1:1. This means that 1 mole of KOH reacts with 1 mole of NiSO4.

Since the mole-to-mole ratio is 1:1, the limiting reactant will be the one with the fewer moles. In this case, NiSO4 has fewer moles (0.0288 moles) compared to KOH (0.044 moles). Therefore, NiSO4 is the limiting reactant.

4. Now, let's calculate the mass of the precipitate (Ni(OH)2) formed using the limiting reactant:
From the balanced equation, we know that the stoichiometric ratio between NiSO4 and Ni(OH)2 is 1:1.

The molar mass of Ni(OH)2 is approximately 92.7 g/mol.

Therefore, the mass of the precipitate formed can be calculated as follows:
mass of Ni(OH)2 = moles of NiSO4 x molar mass of Ni(OH)2 = 0.0288 moles x 92.7 g/mol = 2.652 g

5. Finally, let's calculate the molarity of all ions left in the solution after the reaction:
Since NiSO4 is the limiting reactant, it will be completely consumed. Therefore, there will be no Ni2+ ions left in the solution.

For the KOH solution, the remaining moles can be calculated as follows:
remaining moles of KOH = initial moles of KOH - moles of KOH consumed in reaction
= 0.044 moles - 0.0288 moles (moles consumed in reaction) = 0.0152 moles

To calculate the molarity of KOH ions, divide the remaining moles by the final volume of the solution:
final volume of the solution = volume of KOH solution + volume of NiSO4 solution
= 220 mL + 120 mL = 340 mL = 0.34 L

Molarity of KOH ions = remaining moles of KOH / final volume of the solution
= 0.0152 moles / 0.34 L = 0.0447 M

Therefore, the molarity of KOH ions left in the solution is 0.0447 M.