Answer:
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
Where is the other statements
It should be a capital C with 4 dots, one on each side.
of oxygen at STP would be required to react completely with 38.8g of propane.
<u>Given that :</u>
molar mass of propane = 44 g/mol
mass of propane = 38.8 g
∴ Moles present in 38.8 g of propane = 
= 0.88 mole
<u>applying rule of balanced equations </u>
1 mole of propane = 5 moles of oxygen
0.88 mole of propane = 5 * 0.88 = 4.4 moles of oxygen
Note : volume of 1 mole of oxygen at STP = 
∴Total volume of oxygen required at STP = 22.4 * 4.4 =
Hence we can conclude that the volume of oxygen at STP required to react completely
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Answer:
The answer is c) charged balloon
