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?

Here is an example of a stoichiometry problem. Follow the steps TWICE; once for one reactant and next for the other reactant. You will obtain two answers; the correct one in limiting reagent problems is the smaller value and the reagent producing that value is the limiting reagent. Post your work if you get stuck.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To solve this problem, we need to follow a step-by-step approach.

1. Determine the limiting reactant:

To identify the limiting reactant, we compare the number of moles of each reactant present. The reactant that has the smallest number of moles is the limiting reactant, meaning it will be completely consumed in the reaction, while the excess reactant will be left over.

First, let's calculate the number of moles for each reactant:

For KOH:
moles of KOH = volume (in L) × concentration
moles of KOH = 0.220 L × 0.200 mol/L
moles of KOH = 0.044 mol

For NiSO4:
moles of NiSO4 = volume (in L) × concentration
moles of NiSO4 = 0.120 L × 0.240 mol/L
moles of NiSO4 = 0.0288 mol

Since KOH has fewer moles (0.044 mol) compared to NiSO4 (0.0288 mol), NiSO4 is the limiting reactant.

2. Calculate the mass of precipitate formed:

To determine the mass of the precipitate formed, we need to use the stoichiometry of the balanced chemical equation between KOH and NiSO4 to calculate the number of moles of the precipitate formed (usually called a double displacement reaction). The balanced chemical equation is:

2KOH + NiSO4 → K2SO4 + Ni(OH)2

From the balanced equation, we know that one mole of KOH reacts with one mole of NiSO4 to produce one mole of Ni(OH)2 precipitate.

Number of moles of Ni(OH)2 = moles of NiSO4 (limiting reactant)

moles of Ni(OH)2 = 0.0288 mol

Next, we need to calculate the molar mass of Ni(OH)2:

Molar mass of Ni(OH)2 = atomic mass(Ni) + 2 × atomic mass(O) + 2 × atomic mass(H)
Molar mass of Ni(OH)2 = 58.69 + (2 × 16.00) + (2 × 1.01)
Molar mass of Ni(OH)2 = 92.69 g/mol

Finally, we can calculate the mass of Ni(OH)2 precipitate:

Mass of Ni(OH)2 precipitate = moles of Ni(OH)2 × molar mass of Ni(OH)2
Mass of Ni(OH)2 precipitate = 0.0288 mol × 92.69 g/mol
Mass of Ni(OH)2 precipitate = 2.66 g

Therefore, the mass of the precipitate formed is 2.66 grams.

3. Calculate the molarity of all ions left in the solution:

To determine the molarity of the remaining ions in the solution, we need to find the number of moles of the excess reactant that did not react completely.

Since NiSO4 is the limiting reactant, KOH is in excess. So, we need to find the moles of KOH that didn't react.

moles of KOH remaining = moles of KOH in the original solution - moles of KOH that reacted
moles of KOH remaining = (0.220 L × 0.200 mol/L) - (0.0288 mol)
moles of KOH remaining = 0.044 mol - 0.0288 mol
moles of KOH remaining = 0.0152 mol

Now, we can calculate the molarity of KOH ions (K+ and OH-) and Ni2+ ions in the remaining solution:

For KOH:
Since 1 mole of KOH produces 1 mole of K+ and 1 mole of OH-, the molarity of K+ and OH- in the remaining solution is the same as the molarity of KOH remaining.

Molarity of K+ and OH- = moles of KOH remaining / volume (in L) of the remaining solution
Molarity of K+ and OH- = 0.0152 mol / (0.220 L + 0.120 L)
Molarity of K+ and OH- = 0.0152 mol / 0.340 L
Molarity of K+ and OH- = 0.0447 M

For NiSO4:
Since 1 mole of NiSO4 produces 1 mole of Ni2+ ions, the molarity of Ni2+ ions in the remaining solution is the same as the molarity of NiSO4 remaining.

Molarity of Ni2+ = moles of NiSO4 remaining / volume (in L) of the remaining solution
Molarity of Ni2+ = 0.0288 mol / (0.220 L + 0.120 L)
Molarity of Ni2+ = 0.0288 mol / 0.340 L
Molarity of Ni2+ = 0.0847 M

Therefore, the molarity of K+ and OH- ions left in the solution is 0.0447 M, and the molarity of Ni2+ ions left in the solution is 0.0847 M.