Cu(s) + 4HNO3(aq) --> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l), suppose that you add 5.0 mL of 6 M nitric acid to a sphere of copper metal that weighs 0.58 grams. which reactant is the limiting reagent?

Answer 1

Cu(s) + 4HNO₃(aq) --> Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

One easy way to identify the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

moles Cu(s) = 0.58 g x 1 mol Cu / 63.55 g = 9.13x10^-3 mols Cu

moles HNO3 = 5.0 ml x 1 L / 1000 ml x 6 mol / L = 0.030 mols HNO3

Divide mols Cu by 1: 9.13x10^-3 mols Cu / 1 = 9.13x10^-3

Divide mols HNO3 by 4: 0.030 mols HNO₃ / 4 = 7.5x10^-3

Since 7.5x10^-3 is LESS THAN 9.13x10^-3 , HNO₃ is the LIMITING REACTANT