Calculate ΔG°for the following reaction at 298 K.  3 Cu(s) + 2NO3-(aq) + 8H+(aq) →3 Cu2+ + 2NO(g) + 4H2O(l) ?

For the values of G° (to 1 decimal place), refer to:
https://www.drjez.com/uco/ChemTools/Standard%20Thermodynamic%20Values.pdf
G°[Cu(s)] = 0 kJ/mol
G°[NO₃⁻(aq)] = -111.3 kJ/mol
G°[H⁺(aq)] = 0 kJ/mol
G°[Cu²⁺(aq)] = 65.5 kJ/mol
G°[NO(g)] = 86.6 kJ/mol
G°[H₂O(ℓ)] = -237.2 kJ/mol
(The values of G° may be slightly different from different sources.)

3 Cu(s) + 2 NO₃⁻(aq) + 8 H⁺(aq) → 3 Cu²⁺(aq) + 2 NO(g) + 4 H₂O(ℓ)  ΔG°

ΔG°
= 3 G°[Cu²⁺(aq)] + 2 G°[NO(g)] + 4 G°[H₂O(ℓ)] - 3 G°[Cu(s)] - 2 G°[NO₃⁻(aq)] - 8 G°[H⁺(aq)]
= [3 (65.5) + 2 (86.6) + 4 (-237.2) - 3 (0) - 2 (-111.3) - 8 (0)] kJ
= -356.5 kJ