Volume/Gas Calculation. (1 Viewer)

[davidbarnes]

a) "Calculate the mass of carbon dioxie formed from 30g of glucose (assuming that all glucose has fermented)."

I worked out the mass of carbon dioxide formed to be 14.5g.

b) "What volume will the above mass of carbon dioxide occupy at 25 degrees celcius?".

I have no idea how to do the above (part b) and did not do this in Year 11. Its something to do with standard lab conditions isn't it. So couls somone please show me what formula to use, and how I would use it in the above question it would be most appreciated.

 

[Forbidden.]

a) "Calculate the mass of carbon dioxie formed from 30g of glucose (assuming that all glucose has fermented)."

I worked out the mass of carbon dioxide formed to be 14.5g.

b) "What volume will the above mass of carbon dioxide occupy at 25 degrees celcius?".

I have no idea how to do the above (part b) and did not do this in Year 11. Its something to do with standard lab conditions isn't it. So couls somone please show me what formula to use, and how I would use it in the above question it would be most appreciated.

RE: b)

Interpreting the data from the 2007 HSC Data sheet,
"1 mole of an ideal gas will occupy 24.79L in volume at 100kPa air pressure and 25⁰C"
or
"1 mole of an ideal gas will occupy 22.71L in volume at 100kPa air pressure and 0⁰C"
But according to your answer you got 14.5g of CO₂ which isn't 1 mole and won't occupy 24.79L at 25⁰C.

There are so many variations of the formula I cannot specifically state one.
But it generally goes as Volume occupied = Moles of gas x Molar Volume

Firstly, we must find out how many moles of CO₂ there are in 14.5g
So since 12.01 + 16 + 16 = 44.01 g.mol^-1 (i.e. 1 mole CO₂ is 44.01g), then 14.5/44.01 therefore there are 0.33 moles in 14.5g of CO₂ (rounded to 2 decimal places).

You want to find out how much volume 0.33 moles of gas occupies at 25⁰C so we use 24.79 L (Not 22.71 L because that's for 0⁰C).

So since Volume occupied = Moles of gas x Molar Volume, then:
0.33 x 24.79 = 8.1807 ... L = 8.18 L (2 d.p).

Therefore 14.5g of carbon dioxide will occupy 8.18 L of air at 25⁰C

 

[davidbarnes]

Thanks Forbidden, seems pretty simple after-all.