izzyrawrz
  • izzyrawrz
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Photon336
  • Photon336
similar idea in the other question
Photon336
  • Photon336
at standard temperature and pressure STP, 1 mole of every gas occupies 22.4L
Rushwr
  • Rushwr
Do u still need help? @izzyrawrz

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izzyrawrz
  • izzyrawrz
Yes please @Rushwr
Rushwr
  • Rushwr
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
  • Rushwr
So if 1 moles occupies 22.4 L Then 4 moles occupy 22.4 *4 = 89.6L
izzyrawrz
  • izzyrawrz
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
  • izzyrawrz
Balanced equation is the same one as the first question
Rushwr
  • Rushwr
oops sorry didn't see this one coming mm wait let me check !
Rushwr
  • Rushwr
nop nop not 4l
Rushwr
  • Rushwr
we know mole/stoicheometric ratio = volume ratio \[\frac{ volume of C3H8 }{ Volume of O2 }= \frac{ Moles ofC3H8 }{ Moles of O2 }\]
Rushwr
  • Rushwr
\[\frac{ 0.7 L }{ Volume of O2 } = \frac{ 1 }{ 5}\]
izzyrawrz
  • izzyrawrz
So then it would be 3.5 L of oxygen?
Rushwr
  • Rushwr
\[Volume of O2 = \frac{ 0.7 * 5 }{ 1 } = 3.5 L\]
izzyrawrz
  • izzyrawrz
Oh I know what I did wrong the first time...I rounded :l
Rushwr
  • Rushwr
oh lol okai ! :)
izzyrawrz
  • izzyrawrz
So thank you for your time :)
Rushwr
  • Rushwr
no problem :)

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