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Okay, I'm lost. Help with Chemistry please.

M

MrCodeDude

Lifer
Hexane (C6H14) burns in air to produce CO2 and H20. How many liters of CO2, at STP, would be produced from burning 26.25 grams of hexane?

I think you set up the equation like this:

26.25g x 1 mole CO2
44g CO2

But how do I go from moles to litres?
-- mrcodedude
 
PV = nRT maybe? 🙂 Hah sorry i just had my organic chem final today and i am unable and unwilling to think any more than that.
 
Originally posted by: Sybren
PV = nRT maybe? 🙂 Hah sorry i just had my organic chem final today and i am unable and unwilling to think any more than that.
Click to expand...
We just learned that today, but he says we should have been able to figure it out without that equation! ARG!
-- mrcodedude
 
Originally posted by: agnitrate
Do you have the answer in the back of the book?

I have an answer but it seems kind of high.

-silver
Click to expand...
These are worksheet problems. There are no answers in the book 🙁! Can you show me what you did?
-- mrcodedude
 
if my memory serves me correct, at STP 1 mole of gas is 22.4 liters. you could use pv = nrt to prove that too.
 
Ok. Here's the initial formula.

C6H14 + 0xygen -> C02 + H20 The balanced equation of that is...

2 C6H14 + 19 Oxygen -> 12 C02 + 14 H20

Take your initial 26.25g C6H14 ...

26.25 g C6H14 1 mole C6H14 12 moles CO2 22.4 L CO2
------------------ x ----------------- x --------------- x ------------ = 41.02 L
1 .................. 86 g C6H14 ... 2 moles C6H14 ... 1 mole CO2

That seems kind of high but I'm not sure the g/mol is right for C6H14 or anything. Chem was last year. The number might not seem so high though since I KNOW that the hexane isn't a solid at room temperature so it would probably yield a lot of gas if you had 25g of it.

Check to make sure I didn't do anything stupid.

-silver

 
Originally posted by: agnitrate
Ok. Here's the initial formula.

C6H14 + 0xygen -> C02 + H20 The balanced equation of that is...

2 C6H14 + 19 Oxygen -> 12 C02 + 14 H20

Take your initial 26.25g C6H14 ...

26.25 g C6H14 1 mole C6H14 12 moles CO2 22.4 L CO2
------------------ x --------------------- x ------------------ x --------------- = 41.02 L
1 86 g C6H14 2 moles C6H14 1 mole CO2

That seems kind of high but I'm not sure the g/mol is right for C6H14 or anything. Chem was last year. The number might not seem so high though since I KNOW that the hexane isn't a solid at room temperature so it would probably yield a lot of gas if you had 25g of it.

Check to make sure I didn't do anything stupid.

-silver
Click to expand...

looks right, but at STP, when you plug it into pv = nrt 1 mole will equal 22.4 L, so pv = nrt is correct
 
Convert your grams of hexane to moles of hexane. Know that for every mole of hexane burned, 6 moles of CO2 are produced. Know that there are 22.4 L of an ideal gas at STP in a mole. In your problem, assume CO2 is an ideal gas at STP.

Ryan
 
1. Write the balanced equation.

2C6H14 + 19O2 --> 12CO2 + 14H2O

2. 1 mole of gas at STP = 22.4L

So convert 26.25 grams of hexane into moles:

26.25 / 86 = .305 mole C6H14

3. Use the mole ratio to convert the moles of Hexane to moles of CO2

.305 mole C6H14 * (12 mol CO2 / 2 mol C6H14) = 1.83 mol CO2 * 22.4 = 41 L CO2

If you use PV = nRT... Then (1)(x) = (1.83)(.08206)(273)
x = 41 L CO2


Edit: Damn it... Someone beat me to it already lol oh well...
 
agnitrate-

I checked your work with a back of the envelope calculation, and it looks right. The answer should be approximately 41 L.

Ryan
 
Originally posted by: rgwalt
agnitrate-

I checked your work with a back of the envelope calculation, and it looks right. The answer should be approximately 41 L.

Ryan
Click to expand...

Wow, it just goes to show you that you DO learn things in high school 😛 I'm so proud of me.

-silver
 
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