I don\'t ask a lot of questions for Chegg experts to answer, but if someone coul

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I don't ask a lot of questions for Chegg experts to answer, but if someone could help me out by solving and explaining this multiple part question, it would help me out tremendously!! thank you, in advance!

1. Iron reacts with oxygen to produce iron(II) oxide, as represented by this equatio) Fez03(s) AH°f- 308 K a. Calculate the number of moles of each of the following before the reaction begins DUE 12/A 824 kl mol-1. A 87.2g sample of Fe(s) is mixed with 13.5 L of Odin at 324 atm and . Fe(s) i. 02(g) b. Identify the limiting reactant when the mixture is heated to produce Fe:0(s).Support your answer with calculations. c. Calculate the number of moles of Fe20s(s) produced when the reaction proceeds to completion. d. The standard free energy of formation, AG?, of Fe:0:(s) is -740. kI mol at 298 K i. Calculate the standard entropy of formation, AS t, of Fe20s(s) at 298 K Include units with your answer Which is more responsible for the spontaneity of the formation reaction at 298 K, the standard enthalpy of formation, 11%, or the standard entropy of formation, asm Justify your answer. ii. he value of AH for the Calculate the standard enthalpy of formation, reaction also produces iron(lI) oxide. 2 Fe0(s)+0:(8) Fe:0.(s).T eaction is -280. kl per mole of Fe203(s) formed. 11% of FeO(s).

Explanation / Answer

a)

before reaction

mol of Fe(s)= mass/MW = 87.2/55.8 = 1.5627 mol

mol of O2 = from ideal gas law

n = PV/(RT) = 3.24*13.5/(0.082*308) = 1.7318 mol

B)

limiting reactant:

form ratio is 2:1

therefore

1.5627 mol of Fe requires --> 1/2*1.5627 = 0.78135 mol of O2, which we do have, therefore O2 is in excess

Fe is limiting

c)

mol fo FeO3 possible to produce

2 mol of Fe = 1 mol of Fe2O3

1.5627 mol of Fe --> 1/2*1.5627 = 0.78135 mol of Fe2O3

d)

dG = dH - T*dS

-740 = -824 - 298*dS/1000

dS = (-740+824 )/(-298)*1000

dS = -281.87 JKmol

ii

the spontaneous process is mostly the -dH nature, since dS negtive wont favour reaction

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