Consider these reactions, where m represents a generic metal. 2m(s)+6hcl(aq)⟶2mcl3(aq)+3h2(g)δh1= −556.0 kj hcl(g)⟶hcl(aq) δh2=−74.8 kj h2(g)+cl2(g)⟶2hcl(g) δh3=−1845.0 kj mcl3(s)⟶mcl3(aq) δh4=−342.0 kj use the given information to determine the enthalpy of the reaction 2m(s)+3cl2(g)⟶2mcl3(s)

The enthalpy of formation of will be -5849 kJ.

The given reactions represent:

Enthalpy of reaction of M with HCl = = -556 kJ

Enthalpy of sublimation of HCl = = -74.8 kJ

Enthalpy of HCl formation = = -1845 kJ

Enthalpy of sublimation of = -342.0 kJ

The enthalpy of formation of will be:

=

= -550 + 6 (-74.8) + 3 (-1845) - 2 (342)

= -5849 kJ.

The enthalpy of formation of will be -5849 kJ.

For more information about enthalpy of formation, refer to the link:

link

i'm assuming that the correct answer is indeed the first one/ "a"..

my dad did alot of meth and ace cheeks