lead-ions-can-be-precipitated-using-a-solution-of-bromide-ions-pb2-aq-2-cl-aq-pb

Question

Lead ions can be precipitated using a solution of bromide ions: $$Pb^{2+}(aq) + 2$$ $$Cl^{–}(aq$$) → PbCl_{2}(s)     ΔH°_{f} = –336.0 kJ/molCalculate ΔH for the formation of 5.15 g PbCl_{2}.

Accepted Answer

–6.22 kJ

Answer

–0.516 kJ

Answer

–4.17 kJ

Answer

–10.19 kJ

Lead ions can be precipitated using a solution of bromide ions: Pb2+(aq) + 2 Cl–(aq) → PbCl2(s) ΔH°f = –336.0 kJ/molCalculate ΔH for the formation of 5.15 g PbCl2.

Lead ions can be precipitated using a solution of bromide ions: Pb²⁺(aq) + 2 Cl^–(aq) → PbCl₂(s) ΔH°_f = –336.0 kJ/mol
Calculate ΔH for the formation of 5.15 g PbCl₂.