What mass of precipitate (in g) is formed when 20.5 mL of 0.300 M Ni(NO₃)₂ reacts with 32.5 mL of 0.300 M NaOH in the following chemical reaction? Ni(NO₃)₂(aq) + 2 NaOH(aq) → Ni(OH)₂(s) + 2 NaNO₃(aq)

Answer 1

You would have to convert both those reactants in terms of moles (multiply volume in Liters by Molarity).

You would determine the limiting reagent (the one that makes less product).

Then use that mole amount to determine the moles of the precipitate (NiOH2)

Then you would multiply that value by the molar mass of NiOH2 to get the value into grams.