Question

i dentify the limiting reactant in the following situation and tell how many grams of excess reactant will remain 72.06g of water and 60.05 g of carbon react to form methane and carbon dioxide

Answer 1

Limiting reactant problems are basically about which reactant runs out first.

Answer 2

mmhh

Answer 3

You will need to use stoichiometry to find the amount of product produced with each amount of reactant. Whichever reactant produces less product is limiting.

Answer 4

ok so find the atomic mass first?

Answer 5

Yup.

Answer 6

ok so is it 18.016

Answer 7

for H2O

Answer 8

Right, 18.016g/mol

Answer 9

and multiply that by 72.06?

Answer 10

To convert to moles, 72.06 g of H2O *(1 mol/18.016g)
That cancels out the grams and converts to moles.

Answer 11

i get 3.99 moles

Answer 12

Good. So, how do we find moles of product from here?

Answer 13

so find the samething for the methane

Answer 14

How do we do that when we don't have the mass?

Answer 15

find the mass for it

Answer 16

How?

Answer 17

atomic mass? idk

Answer 18

We write the balanced reaction and use the ratio of the coefficients. This will be the amount of product we can produce from the amount of H2O we have.

Answer 19

From the moles of H2O we found so far.

Answer 20

Can you write the reaction and balance it?

Answer 21

2C+2H2O____CH4+CO2

Answer 22

Awesome, so 3.99 moles of H2O *(1 mol CH4/2 mol H2O) cancels out moles of water and gives moles of methane.

Answer 23

1.995mol CH4

Answer 24

Sweet :) You're almost halfway done!

Answer 25

Can you find the moles of methane produces from the mass of C given?

Answer 26

for the 2 carbons right

Answer 27

so i would do 60.05x 1mol/24.02??

Answer 28

Yup :) That gets you moles of Carbon.

Answer 29

2.5 mol of carbon :)

Answer 30

:) What now?

Answer 31

make a ratio?

Answer 32

Sweet, what ratio would be helpful?

Answer 33

2mol of C/2mol of H2O??

Answer 34

How would that help us?

Answer 35

We are trying to compare the amount of product we can produce from each amount of reactant.

Answer 36

so take the 2.5 mol of C2

Answer 37

2.5 mol of C (Carbon isn't diatomic. We just have 2 moles of C in each reaction.)

Answer 38

ok

Answer 39

So, the ratio we need now is 1 mol CH4/2 mol C

Answer 40

To find the moles of methane from the amount of carbon we started with so we can compare the moles of methane from the amount of water.

Answer 41

oh ok

Answer 42

So, can you tell now which one is limiting?

Answer 43

methane

Answer 44

or CH4

Answer 45

Methane can't be limiting. It is a product. The limiting reactant must be a reactant...

Answer 46

C2

Answer 47

Close. It is C, not C2.

Answer 48

Carbon is not diatomic

Answer 49

ohhh ok

Answer 50

Good! Now all you have to do is find out how much water you would have left over, since it is in excess.

Answer 51

and how will i do that

Answer 52

Now that you know how many moles of CH4 that can be produced from the limiting reactant, you can back calculate the grams of water needed using stoichiometry again. Then grams you started with minus grams used is grams left over.

Answer 53

Let me know how that goes. I have to sign off for a couple of hours. Have fun! :)

Answer 54

ok so 72.06-3.99

Answer 55

which is 68.07 g of excess?

Answer 56

I don't think that is right. I'll be back later. Sorry.

Answer 57

oh well b 4 u go can u atlest tell me what to subtract

Answer 58

so would you do 3mol H2Ox 18.02gh20/1mol n get 54.06g of water as the excess reactant

Answer 59

Since 1.25 mol of CH4 is the amount of methane produced before the carbon runs out, we can back -calculate the amount of H2O reacted by
1.25 mol CH4*(2 mol H2O/1 mol CH4)

Answer 60

Use molar mass to find grams of H2O. This is the mass reacted.

Answer 61

So, mass left over is mass you start with minus mass reacted.

Answer 62

I hope this helps! :)

Answer 63

ok i got 2.5

Answer 64

where did u get 1.25 molCH4 from?

Answer 65

So, we made a mistake earlier and I didn't catch it until now. The molar mass of C is. 12.01, not 24.02.

Answer 66

So, we made a mistake earlier and I didn't catch it until now. The molar mass of C is 12.01, not 24.02.

Answer 67

Can you see what I wrote on this whiteboard? http://172.16.42.7:26290