Question

can someone please help me find these values?

Answer 1

@phi

Answer 2

@robtobey

Answer 3

For someone to comment, he's have to know more about the lab.

Answer 4

the moles you can simply find by dividing the masses by their molar mass... for the volume of H2 formed, you need to write the equation of the reaction in your exp and work with that.

Answer 5

@aaronq so for magnesium it would be 24.31/0.035?

Answer 6

the other way around

Answer 7

oh ok and for h2? @aaronq

Answer 8

you have to find the volume first

Answer 9

is the volume of hydrogen gas produced the same as teh volume of liquid displaced? @aaronq

Answer 10

unless you made other gases in your experiment

Answer 11

yes

Answer 12

no i didnt

Answer 13

so the volume divided by the mass of h2 will give me the moles of h2?

Answer 14

well for ideal gases you can just use the fact that 1 mole = 22.414 L at STP

Answer 15

otherwise you have to know the density of the gas

Answer 16

therefore, how would the equation look?

Answer 17

so for ideal gases,

moles = volume of gas/(22.414 L/mol)

Answer 18

make sure you have the same units

Answer 19

oooh

Answer 20

@aaronq for the moles of magnesium igot 0.001 for all the trials, but my friend wrote it in dimensional analysis . whats betteR?

Answer 21

it's the same thing, since it's a gas, the volume is proportional to the amount of moles

Answer 22

ok and how would i find the ratio of volume displaced in mL to moles h2 gas produced? @aaronq

Answer 23

a ratio is just a division

volume of gas/moles of gas

Answer 24

so volume of h2/moles of h2?

Answer 25

@aaronq

Answer 26

yeah

Answer 27

@aaronq i get a huge number

Answer 28

37.00ml/0.002

Answer 29

yep, thats what the want

Answer 30

18500?

Answer 31

they actually want the volume displaced/molesof H2 which is the same thing

Answer 32

mL/mol

Answer 33

so its right?

Answer 34

@aaronq

Answer 35

to what they're asking you, yes

Answer 36

but its such a big number? @aaronq

Answer 37

yep thats not that big of a number, remember it's in mL ..

Answer 38

18.5 L DAMM

Answer 39

lol that's not a lot, it really should be around 22.414 L

Answer 40

ARE YOU POSITIVE?

Answer 41

well unless they want you to calculate using

Mg (s) + HCl -> H2 + MgCl (aq)

Answer 42

how would that be?

Answer 43

this would be the theoretical way

you balance the equation and use diminutional analysis, but really your answer is gonna be like off by very little, also, because of rounding

Answer 44

did they give you the concentration of the acid?

Answer 45

no

Answer 46

all we had to do was place a piece of magnesium inside a buret that contained hcl and this was then converted and all the hcl migrated down to the bottom of the buret

Answer 47

basically the magnesium was reacting with the hcl

Answer 48

and after that we had to find the values in that table

Answer 49

hm well they really wouldn't need to tell you, do you know what a limiting reagent is?

Answer 50

limiting reactant?

Answer 51

yep, it's the same thing.

you could use the number of moles of that and translate it to moles of H2 produced

2Mg (s) + 2HCl -> H2 + 2MgCl (aq)

0.001439 moles of Mg/2 = x moles of H2/1

moles of H2 = 0.0028794734677088

thats the theoretical yield

Answer 52

well that makes more sense because thats what we are currently learning in class haha

Answer 53

oh wait sorry i multiplied, it should be

moles of H2 = 0.0007198683669272

Answer 54

haha oh true, well maybe let me know next time? haha

Answer 55

sorry!!!

Answer 56

so in other words, to find lets say trial #1

Answer 57

where the volume of hydrgen gas produced was 37.00mL and the moles of h2 gas produced were 0.002 mol

Answer 58

you would convert the mass of Mg to moles

then using the balanced equation, build a relationship between H2 and Mg using their respective coefficients..
this just means:

moles of Mg/coefficient=moles of H2/coefficient

Answer 59

oh my god

Answer 60

moles of mg that reacted?

Answer 61

coefficient is 2?

Answer 62

yeah because thats your limiting reactant..

Answer 63

for Mg is 2 for H2 is 1

Answer 64

wait no i balanced it wrong because it's

Mg + 2HCl -> MgCl2 + H2

Answer 65

so it's coefficient is 1 for both

Answer 66

sorry, i'm like lost today idk why

Answer 67

i dont get it

Answer 68

Mg (s) + 2HCl -> H2 + MgCl2

0.001439 moles of Mg/1 = x moles of H2/1

moles of H2 = 0.001439

thats the theoretical yield

which is close to what you would get if you just divided the volume of gas produced by 22.414 L

Answer 69

thats the ratio?

Answer 70

yep the ratio is 1:1

Answer 71

but what value would i put?

Answer 72

i would go with the theoretical since thats what you're doing in class

Answer 73

0.001439?

Answer 74

yeah, even round up to 0.0015

Answer 75

or down to 0.0014

Answer 76

ok wait where did the 0.001439 come from>?

Answer 77

its the number of moles of magnesium

Answer 78

ooooh yes thats my 0.001

Answer 79

haha yep, i mean round to whatever you think it's best, or whatever they're teaching you at school

Answer 80

but for the ratio isnt moles of h2 divided by moles of magnesium? so 0.002h2/0.001mg

Answer 81

which ratio?

Answer 82

ratio of volume displaced to moles h2 gas prodced

Answer 83

well the amount of moles of H2 produced is equal to the moles of Mg that reacted according to the stoichiometric ratios of the equation...

so to answer your question it would be

37 mL/0.001 moles.. which is what you had before, if i remember correctly

Answer 84

37 being the volume of hydrogen gas produced and 0.001 moles of magnesium reacted?

Answer 85

well, technically yes, but the question is asking for the volume displaced divided by the moles of H2 produced..which are the same values

Answer 86

oh so volume of liquid displaced divided by the moles of h2

Answer 87

and thats the ratio?

Answer 88

wow i feel stupid

Answer 89

that they're asking you for in the question yes.

They're basically seeing how close that value is to the ideal gas constant volume
of 1 mole =22.414 L, since

37mL/0.0014 mol = 26428.5714285714285714 mL/mol

26.428 L/mol


so it's close if you use more than 0.001, which i think you should do lol

Answer 90

oh my i get the big number again hahaha

Answer 91

haha don't feel stupid

Answer 92

your friend in the other thread used 0.00144 moles

Answer 93

0.0014 is in the correct significant figures? i keep using 0.001

Answer 94

hm idk though if you don't use that your error is pretty large

Answer 95

exactly