dimensional analysis?

Question

I need help setting up this problem:

2 C4H10(g) + 13 O2(g) -> 8 CO2(g) + 10H2O(g)

How many liters of H2O(g) is formed when 0.529 L of C4H10(g) is burned?

Answer 1

C4H10 is butane, so

0.529 L of Butane x (1 mol/22.4 L) Butane x (10 mol Butane/2 mol Water) x (22.4 L/1 mol) Water = 2.65 L of Water.

The 22.4 L cancel out, so you really don't need it. 1 mole of gas at STP will always occupy 22.4 L. I think this is your anwser, it has been 10 years since I've done something like this.

Answer 2

You can absolutely do this with volumetric amounts. Your problem is most likely referring to STP (Standard Temp, Pressure) which is 1 atm and 298 K.

There are two ways to do this: the beginner's way, and the advanced way.

The beginners way uses the fact that 1 mol of an ideal gas = 22.414 L

Use the ideal gas law to convert moles to liters. PV = nRT

So, convert your liters of butane into moles.
Compare this number with the number of moles of water produced. (i.e. do the stoichiometry)
Once you have the moles of water produced, convert them to liters using the ideal gas law or the 22.414 conversion factor.

Note: the ideal gas law is appropriate to use when you're given a pressure different than 1 atm, 298 K.

Good luck.

Look up molar volume with google if you need more in-depth example and explanation.

Answer 3

Assuming the gases (butane C4H10 and H2O) are at STP then

0.529L/22.4L = #moles of C4H10

#moles of C4H10 x 10mole H20/2mol C4H10 =# Moles of H2O produced.

#moles of H2O produced x 22.4 L/mole = volume of H2O formed.
Notice that the volume produced in the same ratio as the number of moles of water to butane.

Answer 4

You can't do that with volume amounts... at least i dont think