Stoichiometry and Gas Laws - Please help

  • #1
Here is the example I was given:

------------------------------

Example:

How many liters of hydrogen gas can be produced at 290 K and 2.5 atm if 40.0 g of sodium react with excess water according to the following equations:

2 Na(s) + 2 H2O(l) --> 2 NaOH(l) + H2(g)?

1. Write down the quantities you know.

T - 290 K
P = 2.5 atm
n = find from sodium and balanced equation.
R = 0.821 L atm/K mol
V = ?

2. Set up the problem.

First find how many moles of hydrogen can be produced from the 40.0 g of sodium (equation).

http://osd.flvs.net/webdav/educator_chemistry_v5/module6/imagmod6/6_14_a.gif



Now use PV = nRT to solve.

(2.5 atm)V = (0.870 mol) (0.0821 L atm)/K mol) (290K)

V = 8.3 L

------------------------------

And here are the questions I am having SO much trouble solving. Please help me out with these!

1) C3H8(l) + 5O2(g) ---> 3CO2(g) + 4H2(g)O(g)

In the combustion of 250 g of propane with an excess of hydrogen, what volume of carbon dioxide is produced at 290 K and 1.5 atm. Choose from the following possible answers:

11904.5 L
270 L
90 L
404 L

2) Magnesium burns in oxygen gas to produce magnesium oxide. What mass of magnesium will react with a 0.25 L container of oxygen gas at 80oC and 770 mmHg?

.209 g
.945 g
.418 g
.124 g

---------------------------

It helps that they are multiple choice, but the example doesn't help me much. I don't understand how to solve these. :grumpy:
 
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Answers and Replies

  • #2
GCT
Science Advisor
Homework Helper
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gotta show your work sugarandspice
 
  • #3
1. If you know grams of propane, you can covert to moles of propane, and then convert to moles of CO2 using the balanced equation. So now what are your known quantities?

2. Start by writing the reaction that will occur (make sure you balance it as well). Then, list the known quantities, and see what you can do.
 
  • #4
15
0
Understand?

Full answer:
1) You need to use pV=nRT
from this equation you know:
p=1.5 atm
Since, 1atm = 101325Pa, then 1.5atm would be 151987.5Pa

R=8.314 Jmol-1K-1
T=290K
You need to calculate number of moles (n)
This is the way:
C3H8 has a molecular mass of 44 (mass of carbon (12) * 3 plus mass of hydrogen (1) * 8) 12*3 + 8 = 44
So, M(C3H8) = 44 gmol-1

Now, 44g : 1mol = 250g : Xmol
x = 250*1/44
x = 5.6818 mol

Now, only thing you don't know is V.

pV=nRT

V=nRT\p

V=0.09m3
0.09m3 = 90dm3

From equation you can see that one mol of propane gives 3 moles of CO2.
Then, we will make an equation:
C3H8(l) + 5O2(g) ---> 3CO2(g) + 4H2(g)O(g)
22.4dm3 3*22.4dm3
90.0dm3 x
x = 270dm3

or

22.4 : 3*22.4 = 90.0dm3 : x

x = 3*22.4*90/22.4
x = 270dm3

One dm3 is equal to 1l, so 90dm3 is equal to 270L.
So answer would be 2) 270L

Hope this helps.
 
  • #5
15
0
Second...

Now, first you need to do is to write the equation:

2Mg + O2 --> 2MgO

Now, you will need to turn 770 mmHg to pascales.
This is the way:
760mmHg is 101325Pa that would meen that 770mmHg is 100010Pa.
80oC needs to be turned to kelvines. How? Look:
T = t + 273.15K, that is:
T = 80 + 273.15
T = 353.15K
We have V, so we will again, use pV=nRT
V = 0.25L = 0.25dm3 = 0.00025m3
Just in this case, we will need to calculate n (number of moles).
n = pV/RT
n = 100010*0.00025/8.314*353.15
n = 0.008515mol
Form first equation (2Mg + O2 --> 2MgO) we can see that 2mol of MG reacts with 1mol of O2.

Set up the equation:
2Mg + O2 --> 2MgO
2mol 1mol
x 0.008515mol
---------------------
x = 0.01703mol of Mg

or

2mol Mg : 1 mol O2 = x : 0.008515mol O2
x = 0.01703mol

Now all we need to do is to turn 0.01703mol of Mg to mass of Mg.
Ar(Mg) = 24.5
1mol : 24.5g = 0.01703mol : x
x = 24.5*0.01703/1
x = 0.417235g

So the answer is 3) 0.417235g
----------------------------------------------------------------

If you ever need anything else, write me. I will help you.
You can send me message on software@alphatechplus.com
 
  • #6
Math Is Hard
Staff Emeritus
Science Advisor
Gold Member
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Hi Stamba,
I think it's great that you are trying to help but please review the guidelines.

https://www.physicsforums.com/showthread.php?t=5374
On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.
 
  • #7
Thanks everyone (Samba especially)! I 'completed' the assignment last week - Looks like I got one right. I just guessed though because I couldn't figure out how to find the amount of moles to use in the PV=nRT equation. I'm still pretty fuzzy about that but thankyou so much Samba your step by step helped clear it up a little bit for me!

Sorry, I didn't know I was supposed to show my work. Next time I will, sorry.
 
  • #8
15
0
Math Is Hard said:
Hi Stamba,
I think it's great that you are trying to help but please review the guidelines.

https://www.physicsforums.com/showthread.php?t=5374

Oh:surprised , sorry!
I haven't read this before.
I won't do it again. o:) Promise!
Sorry, about this, all.
 

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