Question

Lead ions can be precipitated from solution with NaCl according to the reaction:

Pb(NO3)2 (aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3 (aq)

Determine the limiting reactant for the reaction between Pb(NO3)2 and NaCl when 0.155L of 1.5M Pb(NO3)2 is mixed with 0.075L of 1.4M NaCl.

Answer 1

Pb(NO3)2 (aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3 (aq)

We see as per stoichiometry, 1 mole of Pb(NO3)2 reacts with 2 moles of NaCl.

Now, in 0.155 L 1.5 M Pb(NO3)2 solution amount of Pb(NO3)2= 0.2325 moles.

And, in 0.075 L 1.4 M NaCl solution amount of NaCl is = 0.105 moles.

0.105 moles of NaCl will react with (0.105/2)=0.0525 moles of Pb(NO3)2.

So, NaCl will be exhausted in the reaction while Pb(NO3)2 will remain excess. NaCl is the limiting reactant in this case.