Question

In the TV show "Breaking Bad" the characters attempt to use HF acid to dissolve guns (among other things). Here we consider instead dissolving guns (which we will assume are pure iron) with sulfuric acid.

Complete the balanced reaction for reacting iron in dilute sulfuric acid to form aqueous FeSO4. Do not worry about formatting subscripts (i.e. O2 to represent diatomic oxygen gas is fine).

Fe + H2SO4 → FeSO4 + __H2(g)___

Answer 1

How many liters of 1 molar sulfuric acid would be required to dissolve 1 kg of iron? Assume the reaction from the previous part goes to completion. The molecular mass of Fe is 55.85 g/mol.

ANSWER: ???????????????????????

It is apparent that they must use an outside source of electrical power to drive the dissolution of the iron. They use a small current so that the dissolution proceeds with the minimum voltage required. Assume standard values for the electrochemical potentials.

How much electrical energy supplied this way is thus required to dissolve an additional 1 kg of iron? Give your answer in kJ.

ANSWER: ????????????????????????????

The characters realize that their hideout has been discovered by the police and they still have a last handgun that weighs 0.25 kg to dissolve. The cops will get there in an hour, so they have to speed up the reaction, by driving it at a higher current. What is the minimum total voltage in volts they'll need to drive the reaction at to get rid of the gun in time? Consider excess potential because of activation losses only and the exchange current I0 to be 1 A for the reaction over the surface of the entire tank (not a current density). α is 0.5 and everything is being done at room temperature. Assume standard electrochemical potentials.

ANSWER: ???????????????????????????

Answer 2

@robtobey

Answer 3

@wio

Answer 4

1st question: 17.9

Answer 5

@wio

Answer 6

@dan815

Answer 7

For the second, use: \(Fe^{2+} + 2 e^- \rightarrow Fe\) \(E_{red}\)= -0.44 V

standard potentials are at 1M of 1 L, so 1 mol, so:

so find moles of Fe (equivalent to 1kg), then multiply by the oxidation potential (+0.44V).


For the 3rd: i'm not sure what \(\alpha\) is, do you know what equation you're supposed to use?

Answer 8

@aaronq can u solve it

Answer 9

nope. you should do this yourself, we're here to help not do it for you.

Answer 10

i got the calculations but i dont got the right one

Answer 11

post what you've done then, we'll check it

Answer 12

@aaronq dont think calculate dcorrectly....

Answer 13

.44 x 1000 x 33

Answer 14

i dont think that is right

Answer 15

units?

Answer 16

if i look at yours it doesn't work but i used two of what u listed but the other one is g

Answer 17

Fe molecular weight is 55.85

Answer 18

can you lay out wher to plug in? the equation?

Answer 19

it should be \(0.44 ~V/mol*n_{Fe}\)

so \(\dfrac{0.44 V/mol*1000g}{55.85 g/mol)}\)

Answer 20

7.878245299910474485228290062667860340197

Answer 21

is that what u listed?

Answer 22

i dont thin it is correct

Answer 23

@aaronq doesn't look right

Answer 24

then you need to find the energy using: V = E/ Q

Q = electron = 1.60217657 × 10^-19 coulombs

Answer 25

so use .44?

Answer 26

electron x .44?

Answer 27

use the value you got after

Answer 28

7.878

Answer 29

so the long answer divided by eletron?

Answer 30

no, you're finding E, so E=V*Q

Answer 31

if u multiply em you get:

0.000000000000000001262234003222918531781557743957027752909586258429

Answer 32

is that what you got?

Answer 33

@aaronq did we get the same answer?

Answer 34

hm its a very small number.. but it looks right.

Answer 35

let me double check

Answer 36

is that in KJ?

Answer 37

no that would be in J

Answer 38

conver tto KJ

Answer 39

times 1000?

Answer 40

divide

Answer 41

if you divide it it become:

0.00000000000000000000126223400322291853178155774395702775291

Answer 42

i think it became smaller....

Answer 43

yeah it gets smaller when you go from a smaller unit to a larger unit

Answer 44

but both answer are wrong....

Answer 45

did you try to calculate it yourself?

Answer 46

to get KJ in the end

Answer 47

the first one is simple stoichiometry:

Fe + H2SO4 → FeSO4 + __H2(g)__

1 kg of Fe to moles

moles of Fe = moles of H2SO4

L of solution= moles of H2SO4/Molarity

Answer 48

i didnt convert Fe to KG.....

Answer 49

omg.....

Answer 50

you're supposed to use it in grams, because the molar mass is in grams per mole

Answer 51

can you map out the equation with the numbers in a better way?

Answer 52

even your setup we get wrong answer

Answer 53

im gona rage quit lol

Answer 54

moles of iron= 1000g/55.845g/mol

\(L_{H_2SO_4}=\dfrac{\dfrac{1000g}{55.845g/mol}}{1~ mol/L}\)

Answer 55

so the answer the previous person gave was right.

Answer 56

wrong.....it says

Answer 57

maybe gotta convert to one decimal format

Answer 58

i dont know anymore

Answer 59

try 18 L

Answer 60

for the answer or for the number to calucaulte?

Answer 61

the answer, just rounded up

Answer 62

wrong

Answer 63

i dont kthink u should round it...

Answer 64

needs a raw answer in kJ

Answer 65

for the first question? it's asking for liters

Answer 66

the second question

Answer 67

someone already gave the first answer " 1st question: 17.9 "

Answer 68

if you read above your post someone answered it already we need the second and third answers

Answer 69

@aaronq we need second and third but you said u dont know third one so we do the second one

Answer 70

i know that. you said it was wrong, so i was checking it

Answer 71

jus raw answer is good enough no round

Answer 72

lol thats not what you said, but whatever.

Answer 73

we just need second question answer

Answer 74

hm it says "an additional kg of iron".. so maybe multiply that number by 2, because we only did it for 1 kg

Answer 75

ok what did u get then?

Answer 76

im not doing the calculations

Answer 77

im already lost...i lost all my numbers

Answer 78

i rage quitted my calculator

Answer 79

go back up, they're somewhere on the page

Answer 80

ok so give me the numebr to calucator x 2

Answer 81

which number do i multiply by 2

Answer 82

the one we found earlier

Answer 83

ill caulaate it jus give me the number

Answer 84

its up in this page, look for it

Answer 85

0.00000000000000000000252446800644583706356311548791405550582 answer is this? can u check ur answer too?i already multipolied it

Answer 86

OMG WRONG!!!!

Answer 87

I RAGE QUIT OMGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

Answer 88

relax man

Answer 89

GONA CUT OF MY HAIR NOW ALL OF IT!

Answer 90

lol

Answer 91

i like few numbers but u give me too many numbers and that has so many zeros i cant handle it

Answer 92

what answer did you get jus post it

Answer 93

need to system check it if it works

Answer 94

2.5*10^-21

Answer 95

wrong

Answer 96

its over....game over man

Answer 97

all the answer give X which is wrong