Assuming that all volume measurements occur at the same temperature and pressure, how many liters of water vapor will be produced in the reaction?
Balanced Equation: C₃H₈ + 5 O₂ - 3 CO₂ + 4 H₂O

Question

Assuming that all volume measurements occur at the same temperature and pressure, how many liters of water vapor will be produced in the reaction?
Balanced Equation: C₃H₈ + 5 O₂ -> 3 CO₂ + 4 H₂O

photon336 · Answer

similar idea in the other question

photon336 · Answer

at standard temperature and pressure STP, 1 mole of every gas occupies 22.4L

rushwr · Answer

Do u still need help? @izzyrawrz

izzyrawrz · Answer

Yes please @Rushwr

rushwr · Answer

yeah so at STP every gas occupies a volume of 22.4 L 
So here water vapour is considered as a gas.
In the reactant side you have 4 moles of Water vapour. If 1 mole occupies 22.4L , 4 moles will occupy 22.4 * 4 will be the volume occupies by 4 moles of water vapor

rushwr · Answer

So if 1 moles occupies 22.4 L 
Then 4 moles occupy 22.4 *4 = 89.6L

izzyrawrz · Answer

Thanks can you help me verify this one
Assuming that all volume measurements occur at the same temperature and pressure, how many liters of oxygen will be required to completely burn 0.700 L of propane gas? 
 I came up with 4 L of oxygen is this correct

izzyrawrz · Answer

Balanced equation is the same one as the first question

rushwr · Answer

oops sorry didn't see this one coming mm wait let me check !

rushwr · Answer

nop nop not 4l

rushwr · Answer

we know mole/stoicheometric ratio = volume ratio
$$\frac{ volume of C3H8 }{ Volume of O2 }= \frac{ Moles ofC3H8 }{ Moles of O2 }$$

rushwr · Answer

$$\frac{ 0.7 L  }{ Volume of O2 } = \frac{ 1 }{ 5}$$

izzyrawrz · Answer

So then it would be 3.5 L of oxygen?

rushwr · Answer

$$Volume of O2 = \frac{ 0.7 * 5 }{ 1 } = 3.5 L$$

izzyrawrz · Answer

Oh I know what I did wrong the first time...I rounded :l

rushwr · Answer

oh lol okai ! :)

izzyrawrz · Answer

So thank you for your time :)

rushwr · Answer

no problem :)