PLEASE!! i need chemistry help, someone please do this problem and it explain it for me!!!?

Question

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4.

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

NiSO4(aq) + KOH(aq) ----->

(b) What precipitate forms?


(c) What is the limiting reactant?


(d) How many grams of this precipitate form?


(e) What is the concentration of each ion that remains in solution?

Ni2+

K+

SO42-

Answer 1

a- NiSO4(aq) + KOH(aq) ----->NiOH(aq)+KSO4(aq)

Answer 2

a) NiSO4 (aq) + 2KOH(aq) ---> Ni(OH)2 (s) + K2SO4 (aq)+ H2O(l)
Simply exchange SO4 with OH (the anions), then balance the charges, and then the eqn.

b) Ni(OH)2

c) 1/10 x 0.2 = 0.02 mol of KOH could be used
1/5 x 0.15 = 0.03 mol of NiSO4 could be used, but you can only use 0.01 mol because there isn't enough KOH, since the ratio is 1:2
Therefore KOH is the limiting reactant. Limiting reactant means the substance that limits the amount of stuff produced.
If there were 0.1 mol of KOH to be used then NiSO4 would be the limiting reactant.

d) 0.1 x Mr of KOH
2 mole of KOH produce 1 mole of Ni(OH)2.
0.2 mole of KOH produce 0.1 mole of Ni(OH)2.

e) I'm not sure about this, sorry.

Answer 3

a) NiSO4 + 2 KOH ---> Ni(OH)2 + K2SO4
b) Nickel hydroxide is the precipitate
c) Given the proportions you have listed, KOH is the limiting reactant (0.02 moles KOH and 0.03 moles NiSO4, and you need 2 moles KOH for every mole of NiSO4).
d) All of your products will be based on the limiting reactant, so you have 0.02 moles of KOH, which will form 0.01 moles of Ni(OH)2. MW Ni(OH)2 is 92.7, so you will have 0.927 g of Ni(OH)2 form.
e) Total volume of solution is now 300 mL. You started with 0.03 moles of NiSO4, where 0.01 moles reacted (of the nickel; the sulfate is a spectator ion). You have 0.02 moles Ni2+ left in solution, which will have concentration of 0.0667 M. Sulfate concentration will be 0.1 M (0.03 moles in 300 mL). Potassium will be 0.0667 M.