correct Question 4 0/2 pts Fe2O3(s) + 2Al(s) Al2O3(s) + 2 Fe(s) AH° = -851.5 kJ/mol...

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correct Question 4 0/2 pts Fe2O3(s) + 2Al(s) Al2O3(s) + 2 Fe(s) AH° = -851.5 kJ/mol If 75 g Fe2O3 is combined with 23 g Al, w

Answer 1

Answer #1

1 mol Fe₂O₃ requires 2 mol Al

moles Fe₂O₃ = 75 g / 159.6882 g/mol = 0.46966526017 mol

moles Al = 23g / 26.98153860 g/mol = 0.85243470882 mol

limiting reactant is Al, so moles Fe₂O₃ required are 0.42621735441 mol

q = 0.42621735441 mol x 851.5 kJ/mol = 362.92407728 ~ 363 kJ

Logic: For

1 mol Fe₂O₃ the value of q is 851.5 kJ, then

0.42621735441 mol Fe₂O₃ (~ 1/2 mol) requires almost half, don't consider Al it is 2 mol.

Answer: 363 kJ

Question 4 0/2 pts Fe2O3(s) + 2Al(s) Al2O3(s) + 2 Fe(s) AH° = -851.5 kJ/mol If 75 g Fe2O3 is combined with 23 g Al, what is t

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Answer 2

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