Find heat of reaction at constant pressure given amounts

Question

When 45.7 mL of 0.388 M lead(II) nitrate reacts with 60.8 mL of 0.543 M sodium chloride, 0.385 kJ of heat is released at constant pressure. What is ΔH° for this reaction?

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

A. –23.3 kJ

B. –4.59 kJ

C. –21.7 kJ

D. –7.58 kJ

E. –43.4 kJ

Answer 1

find moles:

0.0457 litres @ 0.388 mol/litre = 0.0177 moles Pb(NO3)3

0.0608 litres @ 0.543 mol / litre = 0.0330 moles of NaCl

using the ratio: Pb(NO3)2(aq) 2NaCl(aq) → , we see that they needed to add twice the moles of NaCl as they add Pb(NO3)3,

however since they added less than twice, it makes NaCl the limiting reagent

the question is... What is ΔH° for this reaction:

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

which uses 2 moles of NaCl:

? kJ for 2 moles NaCL @ -0.385kJ / 0.0330 moles of NaCl = -23.333

your answer is (3 sig figs): dH = -23.3kJ  "A"