25.0 ml of 0.235 M NiCl2 are combined with 30.0 ml of 2.260 M KOH. How many grams of nickel hydroxide will precipitate?

Answer 1

NiCl2 + 2KOH ==> Ni(OH)2 + 2KCl

mols NiCl2 = M x L = approx 0.006 but you need a more accurate answer.
mols KOH = M x L = approx 0.07--again you need a more accurate answer.

mols Ni(OH)2 formed if all NiCl2 used will be 0.006 x 1 mol (Ni(OH)2/1 mol NiCl2) = approx 0.006

mols Ni(OH)2 formed if all KOH used will be 0.07 x (Ni(OH)2/2KOH) = approx 0.035

You see the answers are not the same; the correct value in limiting reagent (LR) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
g Ni(OH)2 formed = mols Ni(OH)2 x molar mass Ni(OH)2