Given that Delta. G for the reaction below is –957.9 kJ, what is Delta. Gf of H2O?

4NH3(g) + 5O2(g) Right arrow. 4NO(g) + 6H2O(g)

Delta. Gf, NH3 = -16.66 kJ/mol
Delta. Gf, NO = 86.71 kJ/mol

6ΔG°(f) H₂O = -229 Kj/mol

Explanation:

                    4NH₃(g)          +      5O₂(g)       =>        4NO(g)           +     6H₂O(g)

ΔG°(f) 4mol(-16.66Kj/mol) | 5mol(0Kj/mol) || 4mol(+86.71Kj/mol) | 6ΔG°(f) H₂O

Hess's Law

ΔG°(Rxn) = ∑ΔG°(f) Products - ∑ΔG°(f) Reactants

-957.9 Kj = [(4mol(+86.71Kj/mol)) + 6ΔG°(f) H₂O(g)] - [4mol(-16.66Kj/mol) + 5mol(0Kj/mol)]

-957.9 Kj = [4(86.7)Kj + 6ΔG°(f) H₂O] - [4(-16.66)Kj] = 346.84Kj + 6ΔG°(f) H₂O + 66.64Kj

ΔG°(f) H₂O = ((-957.9 - 346.84 -66.64)/6)Kj =  -228.56 Kj ≅ -228.6 Kj*

*Verified with Standard Heat of Formation Table

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2