?! Help!! Part of the process of recovering zinc metal from zinc sulfide ore is

Question

?! Help!! Part of the process of recovering zinc metal from zinc sulfide ore is depicted in the following chemical equation. 2 ZnS(s) + 3 O_2(g) rightarrow 2 ZnO(s) + 2 SO_2(g) What is the work done on this system if 20.0 grams of zinc sulfide are reacted with excess oxygen at 298 K? A) 2.51 J B) 254 J C) 762 J D) 508 J 23. A system at constant pressure produces -5.0 J of heat. What is DeltaH for this system? A) 0.0 J B) it depends on the work performed C) -5.0 J D) 5.0 J 24. The oxidation of iron metal by oxygen produces iron(III) oxide according to the following equation. 4 Fe(s) + 3 O_2(g) 2 Fe_2O_3(s) DeltaHdegree_don = -1648 kJ.mol^-1 Calculate the energy released when this reaction produces 25.0 grams of iron(III) oxide at constant pressure. A) -65.9 kJ B) -129 kJ C) 258 kJ D) -4.12 times 10^4 kJ

Explanation / Answer

22)

work done = P x dV = dn x RT

here

dn = moles of gases in products - moles of gases in reactants

now

moles = mass / molar mass

so

moles of ZnS taken = 20 / 97.474 = 0.205183

now

2 ZnS + 302 --> 2 ZnO (s) + 2 S02 (g)

moles of 02 reacted = 1.5 x moles of ZnS

moles of S02 formed = moles of ZnS

so

dn = moles of ZnS - 1.5 x moles of ZnS

dn = -0.5 x moles of ZnS

dn = -0.5 x 0.205183

dn = -0.1025915

now

W = dn x R x T

W = -0.1025915 x 8.314 x 298

W = -254 J

so

work done on the system is 254 J

the answer is B) 254 J

23)

we know that

dH = dU + d(PV)

dH = dU + PdV + VdP

at constant prssure

dP = 0

so

dH = dU + PdV

dH = dQ

given

dQ = -5

so

dH = -5 kJ

the answer is C) -5 J

24)

we know that

moles = mass / molar mass

so

moles of Fe203 = 25 / 159.69 = 0.15655

consider the given reaction

4 Fe + 302 ---> 2 Fe203 dH= -1648 kJ/mol

2 mole of Fe203 ---> -1648

0.15655 mole of Fe203 ---> y

y = -1648 x 0.15655 / 2

y =-129 kJ

so

energy released is -129 kJ

the answer is B) -129 kJ

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