Question

If 100. g of iron metal are combined with 200. g of an 80% hydrochloric acid, HCl, solution, iron(III) chloride is produced along with a gas. What is the limiting reactant? How many cm3 of the gas will be produced if the percent yield is 70.0%? If the gas is burned in air which is 21% oxygen, what volume of air will be needed for the reaction?

Answer 1

It is a multiple step question, break down the problem to make it easy. go step by step
1) First you have to write down the equation and balance it.
2) Then calculate the theoretical yield for the reaction with the 100g of iron and the theoretical yield with the hydroclhoric acid (assume that this acid is 80% mass/vol) . The one that gives a smaller theoretical yield is the limiting reactant.
3) Calculate the theoretical yield of the gas with the limiting reactant and then calculate the 70% of this value. Assume that the reaction is a STP however problem doesnt said that, to convert the moles (mass) into volume of gas.
4) write down a reaction for the combustion of the gas with oxygen. Calculate how much oxygen you need to convert all the gas into products. Consider the air is only 21% (vol/vol) oxygen and again consider STP conditions and calculate the volume of air that will contain that amount of oxygen.

Answer 2

Thank you!!!!

Answer 3

So I'm having trouble with the theoretical yeild. Can you help

Answer 4

let me know if you need more help, the 80% hydrochloric acid concentration is not well defined in the problem. It is not clear if it is mass solute/mL solution, mL solute/mL of solution, gr solute/gr solution, or any other combination. Looking back to the problem i think that 200 g HCl acid 80% to make the problem easy will be mass of HCL/100 g of solution.

Answer 5

did you write down the equation and balanced?

Answer 6

Yeah. 3Fe+HCl=Fe3Cl+H

Answer 7

the H is a molecular element is never H it is as a gas H2 (g)

Answer 8

do you know about the molecular elements and the rule of 7?

Answer 9

No

Answer 10

the equation is not correct, the products are wrong (both of them) check the formula again , the H has to be H2 and the iron (III) chloride formula also is wrong

Answer 11

That's a direct copy from the homework sheet

Answer 12

Oh wait you ment my answer

Answer 13

3Fe+HCl=Fe3Cl+H (wrong!!!)

Answer 14

Ah

Answer 15

What would it be then?

Answer 16

Can you explain the question and give me the answer. Being able to see the answer will help to understand the problem

Answer 17

It's Fe + 2HCl = FeCl2 + H2

Answer 18

no

Answer 19

To what?

Answer 20

2Fe + 6HCl = 2FeCl3 + 3H2

Answer 21

Man I was way off. Thanks. So what next?

Answer 22

now find in the periodic table the atomic mass of Fe, H, Cl and calculate the molecular mass of the HCl and H2

Answer 23

92 combined. Or did you want HCl is 36 and Fe is 56

Answer 24

And H2 is 2

Answer 25

that is OK,
calculate the theoretical yield of H2 according to the stoichiometry of the equation

mass of Fe -> moles of Fe -> moles of H2
100 g Fe x (1 mol Fe/ 56g Fe) x (3mol H2/ 2 mol Fe) = ?

mass HCl -> % mass -> mol HCl -> mol H2
200 g HCl x (80/100) x (1molHCl/36g HCl) x (3mol H2/6 mol HCl) = ?

do these two calculation and find out which one gives you the smaller number, that is going to be your limiting reactanct

Answer 26

Fe= 300 and HCl is 160
so HCl is limiting?

Answer 27

300 what?

Answer 28

grams

Answer 29

no, 2.678 mol H2= 5.356g H2

Answer 30

i just got that, yeah. I did the problem weird

Answer 31

and the HCl?

Answer 32

2.22 g

Answer 33

HCl is still the limiting

Answer 34

these are H2 moles

Answer 35

yes the limiting is HCl acid!!!!!

Answer 36

yay! ok so what next?

Answer 37

because the reaction is not 100% efficient but only 70% you have to calculate the 70% of your theoretical yield with the limiting reactant (2.22 mol H2) to find out how much H2 you get in the real yield.
In few words calculate the 70% of 2.22 mol H2

Answer 38

1.554g

Answer 39

or is that mol

Answer 40

moles!!!

Answer 41

ok so now that we have that, what do we do?

Answer 42

now the problem is asking you to convert that moles of H2 to cm3. It doesnt said but I will assume that the gas is in STP conditions (1 atm and 273K). At that conditions the volume of the one mole of any gas will be 22.4L. So you can multiply the 1.554 mol H2 times the conversion factor (1mol / 22.4L) and you will get the volume in L, then convert L to cm3 (1000 cm3/1L)

Answer 43

would the volume be 34.84L

Answer 44

i rounded the mole to 1.6,

Answer 45

yea something like that now convert to cm3

Answer 46

now you have to calculate the volume of air if you burn that H2 with air with a 21% of O2

Answer 47

so the cm^3 would be 35,840 cm^3

Answer 48

yea!

Answer 49

cool! so now how do we do the next part

Answer 50

you have to write down a new equation

Answer 51

what is that equation?

Answer 52

hydrogen + oxygen -> products ?

Answer 53

h2o2?

Answer 54

because it oxygen gas right?

Answer 55

yes but I will think that after the "combustion" you will get H2O than H2O2

Answer 56

oh ok so now what do i do with that

Answer 57

write the reaction

Answer 58

I'm confused, isn't h2o the reaction?

Answer 59

H2O will be the product of your reaction, but you have to write the chemical equation of H2 + O2 and balance the equation to figure out how much O2 you will need to burn all the H2 that you got in the previous reaction

Answer 60

It is this your first chemistry class ever?

Answer 61

yeah

Answer 62

the teacher isn't very good, so we have to learn a lot on our own

Answer 63

Let the teacher know that the problem is a little complicated and is missing some of the detail information such as when you use a % concentration is confusing if doesnt said g/L or g% ,etc, and when she ask about the volume of the gas has to said what are the pressure and temperature. You well know that a gas a different pressure or temperature is going to have different volume? then the answer for that part of the problem could be any volume depending of the conditions of pressure and temperature

Answer 64

its sounds like it is g/L. We answered the first part, all thats left is the oxygen combustion. What do you think that would be if you had to guess

Answer 65

i really need to get this

Answer 66

I will said by volume, L/L, that means 21L of O2 every 100L of dry air

Answer 67

write down the equation

2H2 + O2 -> 2 H2O

Answer 68

ok

Answer 69

the equation is telling you that for each 2 moles of H2 you will need 1 mole of O2

Answer 70

ok, so we need to find the volume of the 21% oxygen?

Answer 71

so if we have 1.554 mol H2 we will need 0.777 moles of O2

Answer 72

is that the answer then?

Answer 73

0.777 moles of O2 at STP conditions will be 0.777 mol x 22.4L = 17.4048 L

Answer 74

then because the O2 is only the 21% of the air, this 17.4048 L is only the 21% of the total air that we need

Answer 75

17.4048L x 100/21 =82.88L of air

Answer 76

ok cool, thats the answer right?

Answer 77

yes at STP conditions!!! if the temperature or the pressure are different you have to use the equation of the gases to convert
P1V1/T1 = P2V2/T2

Answer 78

ok thank you so much!!! your the best!

Answer 79

any other problem?

Answer 80

no thats it, but if i have anyone question to get help with ill message you!