Asolution 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. (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?

a) 2KOH + NiSO₄ → Ni(OH)₂ + K₂SO₄

b) The precipitate that forms is Ni(OH)₂

c) We calculate the moles of each reagent:

0.1 L * 0.200 M KOH = 0.02 mol KOH

0.2 L * 0.150 M NiSO₄ = 0.03 mol NiSO₄

The amount of KOH needs to be twice the amount of NiSO₄, so for 0.03 mol NiSO₄ it would take 0.06 mol KOH. There are not as many, so KOH is the limiting reactant.

d) We calculate the amount of product formed using the limiting reactant:

0.02 mol KOH 1.8538 g Ni(OH)₂.

e) The final volume is 0.3 L

[K⁺] = 0.02 mol / 0.3 L = 0.067 M

[SO₄⁻²] = 0.03 mol / 0.3 L = 0.100 M

The Ni⁺² moles that reacted were: 0.02 mol KOH * = 0.01 mol Ni⁺²

[Ni⁺²] = 0.02 mol / 0.3 L = 0.067 M

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.