Question

MEDALS WILL BE REWARDED!
The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required to completely react with 0.025 g C3H8?

0.018
0.034
0.045
0.091

Answer 1

@whpalmer4 @SolomonZelman
Sorry to bother you guys, but I saw that y'all are online and studying chemistry and was wondering if y'all could help me. ?

Answer 2

I suck at science, sorry :(

Answer 3

It's fine. Thank you for replying though. :)

Answer 4

First, don't we need to know the reaction?

Answer 5

If I am completely honest, I don't understand this at all.

Answer 6

No problem. We're burning some propane, right?

Answer 7

I do believe so

Answer 8

We need to make a balanced equation that uses up all of the atoms on the left producing the stuff on the right.

nC3H8 + mO2 = C3n + H8n + O2m
as a first step

Answer 9

if we let n = 1, so we are reacting 1 propane molecule with an unknown number of oxygen molecules, that gives us

C3H8 + mO2 = 3C + 8H + O2m

Now the combustion products are going to be what? What do you get when you burn a hydrocarbon like propane in plenty of oxygen?

Answer 10

You get carbon dioxide and water, right? CO2 and H2O

Answer 11

so if we stick with our initial guess of n = 1, we've got 3 carbon atoms to pair up with oxygen. How many oxygen atoms will we need to make the carbon atoms worth of CO2?

Answer 12

6 atoms of oxygen

Answer 13

Good. That's 3 molecules of O2.

C3H8 + 3O2 = 3CO2 + H8

uh, but wait, we've got a balanced equation, but we didn't react all that hydrogen. What are we going to do about that?

Answer 14

The hydrogen and the other oxygen atoms will combine to make H2O

Answer 15

What other oxygen atoms?

Answer 16

We need to put them in the equation. Can you do so?

Answer 17

oops sorry I read that wrong, and I'll try.

Answer 18

Um...
C3H8 + 3O2 = 3CO2 +4H2 ?

Answer 19

I thought we were going to make some H2O?

Answer 20

If we have already used our oxygen atoms to create 3CO2, what oxygen do we have left to combine with the hydrogen to make any H2O?

Answer 21

@whpalmer4
Hello? Sorry not trying to be pushy.

Answer 22

Well, we need to add some more oxygen to the equation! How much more oxygen do we need to use up 8 H atoms?

Answer 23

We need 4 oxygen atoms to cancel out all the hydrogen atoms when making H2O

Answer 24

and how many oxygen molecules will that be?

Answer 25

2

Answer 26

so 2O2

Answer 27

Right. So what is our final equation?

Answer 28

(sorry, my kid is asking me for help with his math homework at the same time, and he naturally gets priority!)

Answer 29

C3H8 + 5O2 = 3CO2 + 4H20
(and don't worry about it. I know what that's like.)

Answer 30

Yes, that looks good to me. Now for the harder part :-)

Answer 31

woohoo (sarcasm)

Answer 32

Any questions about how we balanced it, before we go on?

Answer 33

I think I'm good.

Answer 34

Good, well, feel free to tag me if you get stuck in the future.

I'll copy part of the problem here so I don't have to scroll back:

The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required to completely react with 0.025 g C3H8?

How many moles of propane are we reacting?

Answer 35

Thank you. :)
and one mole

Answer 36

Okay, are you sure about that? The molar mass of C3H8 is 44.1 g/mol, which to me suggests that 1 mol of C3H8 weighs in at 44.1 g, but we are only reacting 0.025 g of C3H8...

Answer 37

Oh yeah...in that case, 1.1025 if I did that right...

Answer 38

Let's think about that for a minute. If I tell you it costs $44.10 to buy a concert ticket, and you check your pockets and tell me that you only have 2 and half cents, do you think you have enough for more than one ticket?!?

Answer 39

\[\frac{0.025\text{ g}}{44.1\text{ g/mol} }= \text{______ mol}\]

Answer 40

0.00057
I did this one first, but the answer didn't seem right. On the other hand, once you used the ticket analogy, it did.

Answer 41

Yes, that's the right number of moles of C3H8. Now, how many moles of O2 are we going to use?

Answer 42

The coefficient in front of C3H8 is missing, so we assume it is 1, given that we have a balanced equation. The coefficient in front of O2 is 5, so we have 5 moles of O2 for every 1 mole of C3H8.

Answer 43

5 x 32 = 160
160 x 0.0005668934240362812 (or 0.00057) =0.09070294784580499 (or 0.091)
so 0.091

Answer 44

Yes, that's the answer I got. I would have set it up as

\[\frac{0.025\text{ g}}{44.1\text{ g/mol}}*5*32\text{ g/mol} = \text{________ g} \]

Answer 45

Yay! and by the way I realize that I do have a question about balancing the equation: How did you know how many atoms should combine with the hydrogen and with the carbon?

Answer 46

well, we know how many carbon and hydrogen atoms we start with, right?

Answer 47

8 hydrogen and 3 carbon

Answer 48

oh so you just fill in as many oxygen atoms as you can with each of the other elements until there are no more O atoms left?

Answer 49

Okay, and we know that we're producing CO2 and H2O, so we do enough O2 to soak up all of the C, and then enough to soak up all of the H

Answer 50

Now sometimes (I can't think of an exact example off the top of my head, unfortunately) you'll end up discovering that you need to add another molecule of the C3H8 (or whatever it is) to get things to balance. That's why I wrote the business with n and m way back when, because I was thinking that this would be such a case.

Say we had

C3H7 + O2 -> CO2 + H2O

if we start the same way, we do 1 C3H7 plus 3 O2 gives us
C3H7 + 3O2 -> 3CO2 + 7H
Now to use up the 7H, we're kind of stuck, because they get used 2 at a time. So we multiply all the quantities by 2:

2C3H7 + 6O2 -> 6CO2 + 14H
now we need 7 O, but that's still not going to work, because O comes as O2, so we double again:

4C3H7 + 12O2 -> 12CO2 + 28H
now we can balance that if we add 7O2 to both sides
4C3H7 + 19O2 ->12CO2 + 14H2O

left side has 4 *3C = 12C
right side has 12C
left side has 4*7H = 28H
right side has 14*H2 = 28H
left side has 19*2O = 38O
right side has 12*2O + 14*o = 24+14 O = 38O

everything balances

Answer 51

I don't think C3H7 exists, but never mind that :-)

Answer 52

Um I'm a little confused, but I think I got it. Also, I cannot thank you enough for all the help you've given me. ^_^

Answer 53

You're welcome! I hope some of it sinks in and helps you through the next problem like this. I know how easy it is to think "oh, yeah, that all makes sense" only to be faced with a blank wall as you stare at the sheet of paper trying to figure out how to do it on your own!