Limiting Reactants

Question

If 3.00 grams of copper(II)nitrate reacts with 4.35 grams of sodium hydroxide, what is the limiting reactant?

1. a. sodium nitrate
2. b. sodium hydroxide
3. c. copper (II) hydroxide
4. d. copper (II) nitrate

Answer 1

First, balance the reaction: Cu(NO₃)₂ + 2 NaOH --> Cu(OH)₂ + 2 NaNO₃. Then do stoichiometry, choosing one of the products (I chose the second as being easier to type!):

(3.00g Cu(NO₃)₂) * (1 mol Cu(NO₃)₂ /187.6g Cu(NO₃)₂ ) * (2 mol NaNO₃ /1 mol Cu(NO₃)₂ ) = 0.0320 mol NaNO₃.

(4.35g NaOH) * (1 mol NaOH /40.0g NaOH) * (2 mol NaNO₃ / 2 mol NaOH) = 0.109 mol NaOH.

So limiting reactant is Copper (II) Nitrate, because there is not enough of it to match the result from NaOH.