When 13.8 mL of 0.870 M lead(II)nitrate reacts with 90.0 mL of 0.777 M sodium chloride,0.279 kJ of heat is released at constant pressure.What is ΔH° for this reaction? Pb(NO₃)₂(aq)+ 2NaCl(aq)→ PbCl₂(s)+ 2NaNO₃(aq)

Question

Quizplus · Accepted Answer

The reaction releases 0.279 kJ of heat for the given amounts of reactants. To find ΔH°, calculate the moles of the limiting reactant (Pb(NO3)2), which is 0.012 moles. Then, divide the heat released by the moles of the limiting reactant: 0.279 kJ / 0.012 mol = 23.3 kJ/mol.

When 138 ML of 0