Explanation:
T = 409.5 K, P = 1.50 atm: V = 22.4 L The ideal gas law is: PV = nRT where. P = pressure. V = volume n = number of moles.
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1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
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Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.