2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)ΔH1=−783.0 kJ2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)ΔH1=−783.0 kJ HCl(g)⟶HCl(aq)  ΔH2=−74.8 kJHCl(g)⟶HCl(aq)  ΔH2=−74.8 kJ H2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJH2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJ MCl3(s)⟶MCl3(aq)  ΔH4=−124.0 kJMCl3(s)⟶MCl3(aq)  ΔH4=−124.0 kJ Use the given information to determine the enthalpy of the reaction 2M(s)+3Cl2(g)⟶2MCl3(s

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