Consider these reactions, where m represents a generic metal.

a. 2 m ( s ) + 6 hcl ( aq ) ⟶ 2 mcl 3 ( aq ) + 3 h 2 ( g ) δ h 1 = − 805.0 k j
b. hcl ( g ) ⟶ hcl ( aq ) δ h 2 = − 74.8 k j
c. h 2 ( g ) + cl 2 ( g ) ⟶ 2 hcl ( g ) δ h 3 = − 1845.0 k j
d. mcl 3 ( s ) ⟶ mcl 3 ( aq ) δ h 4 = − 484.0 k j
use the given information to determine the enthalpy of the reaction
2m(s)+3cl2(g) --> 2mcl3(s)h=

-5820,8 kJ

Explanation:

It is possible to sum ΔH of half-reactions to obtain the ΔH of a global reaction (Hess's law).

Using the reactions:

a. 2M(s) + 6HCl(aq) ⟶ 2MCl₃(aq) + 3H₂(g) ΔH₁ = −805.0 kJ

b. HCl(g) ⟶ HCl(aq) ΔH₂= −74.8 kJ

c. H₂(g) + Cl₂(g) ⟶ 2HCl (g) ΔH₃ = − 1845.0 kJ

d. MCl₃(s) ⟶ MCl₃(aq) ΔH₄= −484.0 kJ

The sum of a + 6b + 3c gives:

a. 2M(s) + 6HCl(aq) ⟶ 2MCl₃(aq) + 3H₂(g) ΔH₁ = −805.0 kJ

6b. 6HCl(g) ⟶ 6HCl(aq) ΔH= 6×−74.8 kJ  = -448,8 kJ

3c. 3H₂(g) + 3Cl₂(g) ⟶ 6HCl (g) ΔH = 3×−1845.0 kJ = -5535,0 kJ

a + 6b + 3c. 2M(s) + 3Cl₂(g) ⟶ 2MCl₃(aq)

ΔH = -805,0kJ - 448,8 KJ - 5535 kJ = -6788,8 kJ

Now, the sum of this reaction - 2d gives:

a + 6b + 3c. 2M(s) + 3Cl₂(g) ⟶ 2MCl₃(aq) ΔH = -6788,8kJ

-2d 2MCl₃(aq) ⟶ 2MCl₃(s) ΔH= 2×+484.0 kJ

The global reaction gives:

2M(s) + 3Cl₂(g) ⟶ 2MCl₃(s)

ΔH = -6788,8kJ + 2×484.0 kJ = -5820,8 kJ

I hope it helps!

the answer is na (sodium)