A solution of .12 L of 0.160 M KOH is mixed with a solution of .3 L of 0.230 M NiSO4.

the equation for this reaction is:
2KOH (aq)+NiSO4 (aq) ----->K2SO4 (aq)+Ni(OH)2 (s)

.89 grams of Ni(OH)2 precipitate form

I need to know the concentration remaining in the solution of:
1) Ni (II)
2) SO4
3) K

I do not know how to do these nor can i find aid to show me how to do it step by step, so to whoever answers this question, could you also show steps so i can learn how to do this? thank you to whoever puts in their time and helps me!!

The first part of this is a limiting reagent problem. I solve these problems by solving two stoichiometry problems. The first one using reagent 1 and all of the other needed; the second time with reagent 2 and all of the other needed. You will obtain two answers; obviously both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is a step by step procedure for solving stoichiometry problems.

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

To find the concentration remaining in the solution of Ni (II), SO4, and K, we first need to determine the moles of each reactant and product involved in the reaction. Then, using stoichiometry, we can find the concentration based on the initial volume of the solutions.

Let's start with the given information and break down the steps:

Given:
Volume of KOH solution = 0.12 L
Concentration of KOH solution = 0.160 M

Volume of NiSO4 solution = 0.3 L
Concentration of NiSO4 solution = 0.230 M

Mass of Ni(OH)2 precipitate formed = 0.89 grams

Step 1: Calculate the moles of KOH and NiSO4:
Moles of KOH = concentration x volume
Moles of KOH = 0.160 M x 0.12 L
Moles of KOH = 0.0192 moles

Moles of NiSO4 = concentration x volume
Moles of NiSO4 = 0.230 M x 0.3 L
Moles of NiSO4 = 0.069 moles

Step 2: Determine the limiting reactant:
To find the limiting reactant, we compare the stoichiometric coefficients of KOH and NiSO4 in the balanced equation. The reactant with the lower stoichiometric coefficient will be the limiting reactant.

From the balanced equation: 2 moles of KOH react with 1 mole of NiSO4

Moles of KOH to react with NiSO4 = 2 x moles of NiSO4
Moles of KOH to react with NiSO4 = 2 x 0.069 moles
Moles of KOH to react with NiSO4 = 0.138 moles

Since 0.138 moles of KOH required to react with NiSO4 is greater than the 0.0192 moles of KOH given, KOH is the limiting reactant. This means all the KOH will react, and we'll have a given amount of NiSO4 in excess.

Step 3: Calculate the moles of Ni(OH)2 formed:
From the balanced equation, we know that 2 moles of KOH react to form 1 mole of Ni(OH)2.

Moles of Ni(OH)2 = 0.5 x moles of KOH
Moles of Ni(OH)2 = 0.5 x 0.0192 moles
Moles of Ni(OH)2 = 0.0096 moles

Step 4: Calculate the remaining moles of Ni (II), SO4, and K:
The moles of Ni (II) and SO4 remaining are related to the moles of Ni(OH)2 formed. Since there is a 1:1 stoichiometric ratio between Ni(OH)2 and both Ni (II) and SO4, the remaining moles of each compound will be the moles of Ni(OH)2 subtracted from the initial moles of Ni and SO4.

Moles of Ni (II) remaining = Moles of NiSO4 - Moles of Ni(OH)2
Moles of Ni (II) remaining = 0.069 moles - 0.0096 moles
Moles of Ni (II) remaining = 0.0594 moles

Moles of SO4 remaining = Moles of NiSO4 - Moles of Ni(OH)2
Moles of SO4 remaining = 0.069 moles - 0.0096 moles
Moles of SO4 remaining = 0.0594 moles

Step 5: Calculate the concentration of Ni (II), SO4, and K:
The concentration is calculated by dividing the remaining moles by the final volume, which is the sum of the initial volumes of the KOH and NiSO4 solutions.

Volume of final solution = Volume of KOH solution + Volume of NiSO4 solution
Volume of final solution = 0.12 L + 0.3 L
Volume of final solution = 0.42 L

Concentration of Ni (II) = Moles of Ni (II) remaining / Volume of final solution
Concentration of Ni (II) = 0.0594 moles / 0.42 L
Concentration of Ni (II) = 0.1414 M

Concentration of SO4 = Moles of SO4 remaining / Volume of final solution
Concentration of SO4 = 0.0594 moles / 0.42 L
Concentration of SO4 = 0.1414 M

Concentration of K = Moles of KOH remaining / Volume of final solution
Since KOH is the limiting reactant, all of it reacts, and there is none remaining. Therefore, the concentration of K is 0 M.

Therefore, the concentrations remaining in the solution are as follows:
1) Concentration of Ni (II) = 0.1414 M
2) Concentration of SO4 = 0.1414 M
3) Concentration of K = 0 M