Calculate the enthalpy change for the reaction mn3o4(s)+co(g)⟶3mno(s)+co2(g) from the following:

mn3o4(s)+4co(g)⟶3mn(s)+4co2(g) δh=255.6kj

mno(s)+co(g)⟶mn(s)+co2(g) δh=102.1kj

express your answer using one decimal place and include the appropriate units.

- 50.7 kJ.

Explanation:

Mn₃O₄(s) + CO(g) ⟶ 3MnO(s) + CO₂(g).

The first reaction be as it is:

Mn₃O₄(s) + 4CO(g) ⟶ 3Mn(s) + 4CO₂(g), ΔH₁ = 255.6 kJ.

The second reaction should be reversed and multiplied by 3 and also the value of its ΔH must multiplied by (- 3):

3Mn(s) + 3CO₂(g) ⟶ 3MnO(s) + 3CO(g), ΔH₂ = (- 3)(102.1 kJ) = - 306.3 kJ.

Mn₃O₄(s) + CO(g) ⟶ 3MnO(s) + CO₂(g).

∴ ΔH rxn = ΔH₁ + ΔH₂ = (255.6 kJ) + (- 306.3 kJ) = - 50.7 kJ.

the answer is c. nacl