pls help me out, do it in details with formula

A sample of 50 cm3 of ethanol gas is burned completely in 200 cm3 of oxygen.
C2H5OH(g) + 3O2(g) -> 2CO2(g) + 3H2O(g)
All volumes are measured at a temperature of 400 K and 1 atm pressure.
What is the total volume of gas when the reaction is complete?
• A 150 cm
• B 200 cm
• C 250 cm
• D 300 cm

PV = nRT
Pressure in Pa, Volume in m^3
1atm = 101000 Pa
cm^3 x 10^-6 -> m^3
For ethanol, n = (101000 x 50 x 10^-6)/8.314 x 400 = 0.0015 moles
For oxygen, n = (101000 x 200 x 10^-6)/8.314 x 400 = 0.006 moles
0.0015 moles ethanol needs 0.0045 moles oxygen, but have 0.006 moles oxygen, oxygen in excess
ethanol + 3 oxygen -> 2 carbon dioxide + 3 water
initial: 0.0015, 0.006, _, _
change: -0.0015, -0.0045, +0.0030, +0.0045
final: 0, 0.0015, 0.0030, 0.0045
total moles = 0.009 moles
Volume = (0.009 x 8.314 x 400)/101000 = (2.963 x 10^-4)m^3 x 10^6 = 296 cm3 = 300 cm3

As it is a multiple choice question, Im sure there is a faster way
Here is the faster way https://www.thestudentroom.co.uk/showthread.php?t=7012791

‘For ethanol and O2 to react in a 1:3 ratio (as in the equation), 150cm3 of O2 will be used - leaving 50cm3 unreacted because it is in excess.

The products will take up 250 cm3 - (50cm3 of ethanol x 5), but the question asks for the total volume of gas, not just the volume of products, so there will be 300 cm3 in total.

Hope this helps lol’

Hi @zunainahaque,

Thank you for posting in the Chemistry Forum.

However, please be aware that generally we do not assist those who ask for full solutions. This is because we want to encourage learning and problem-solving rather than simply providing answers.

I'll keep the solution provided above for now, but in the future, could you please share your own attempt first? This way, we can offer personalised assistance.

We appreciate your cooperation and look forward to seeing you continue to contribute to the Chemistry Forum!


5hyl33n

Hi @zunainahaque,

Thank you for posting in the Chemistry Forum.

However, please be aware that generally we do not assist those who ask for full solutions. This is because we want to encourage learning and problem-solving rather than simply providing answers.

I'll keep the solution provided above for now, but in the future, could you please share your own attempt first? This way, we can offer personalised assistance.

We appreciate your cooperation and look forward to seeing you continue to contribute to the Chemistry Forum!


5hyl33n

thank you very much for this. i already got my answer.

you are always welcome to delete the solution, imagine im blind and delete it like i never saw it. go ahead, please!

Hi @zunainahaque,

Thank you for posting in the Chemistry Forum.

However, please be aware that generally we do not assist those who ask for full solutions. This is because we want to encourage learning and problem-solving rather than simply providing answers.

I'll keep the solution provided above for now, but in the future, could you please share your own attempt first? This way, we can offer personalised assistance.

We appreciate your cooperation and look forward to seeing you continue to contribute to the Chemistry Forum!


5hyl33n