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1. Can anyone help me with gas calculations? I'm struggling with trying to convert moles to mass in relation to gases, do you just do the normal m=NXM. I was trying to do this for question d in the analysis section and I got lost..

http://www.ocr.org.uk/Images/67673-u...k-specimen.pdf
2. (Original post by kath_igidbashian)
Can anyone help me with gas calculations? I'm struggling with trying to convert moles to mass in relation to gases, do you just do the normal m=NXM. I was trying to do this for question d in the analysis section and I got lost..

http://www.ocr.org.uk/Images/67673-u...k-specimen.pdf
Pure substances:

moles = mass/Mr

Solutions:

moles = concentration x volume

Gases:

moles = mass/22.4 (at STP)

BUT, don't forget that gases can also be pure substances, in which case equation 1 also applies.
3. (Original post by charco)
Pure substances:

moles = mass/Mr

Solutions:

moles = concentration x volume

Gases:

moles = mass/22.4 (at STP)

BUT, don't forget that gases can also be pure substances, in which case equation 1 also applies.
Thanks! Is there anyway you could take a look at the question I said and do a work through guide for me? Just to show me the steps in case I get another question like this again?
4. (Original post by kath_igidbashian)
Thanks! Is there anyway you could take a look at the question I said and do a work through guide for me? Just to show me the steps in case I get another question like this again?
part (d) comes from the answers to (a), (b) and (c).

What have you got for those?
5. (Original post by charco)
part (d) comes from the answers to (a), (b) and (c).

What have you got for those?
MgCO3 is 0.00356 mol
H2SO4 is 0.0005 mol

H2SO4 is in excess
6. (Original post by kath_igidbashian)
MgCO3 is 0.00356 mol
H2SO4 is 0.0005 mol

H2SO4 is in excess
what about the equation for the reaction?
7. (Original post by charco)
what about the equation for the reaction?
oh sorry
H2SO4(aq) + MgCO3 (s)-> MgSO4(aq) + H20(l) + CO2(g)
8. (Original post by kath_igidbashian)
oh sorry
H2SO4(aq) + MgCO3 (s)-> MgSO4(aq) + H20(l) + CO2(g)
OK, so the equattion tells you that the moles of carbon dioxide will be equal to the moles of the limiting reagent.

MgCO3 is 0.00356 mol
H2SO4 is 0.0005 mol

which wuold make magnesium carbonate in excess ...

But 0.1 x 0.05 moles of sulphuric acid = 0.005 mol NOT 0.0005 mol
0.3 g of magnesium carbonate = 0.3/84 = 0.00357 mol

So, yes sulphuric acid is in excess and magnesium carbonate is the limiting reagent.

Hence moles of CO2 = moles of magnesium carbonate = 0.00357 mol.

1 mol of gas at RTP = 24000 cm3

hence 0.00357 mol = 0.00357 * 24000 = 85.7 cm3
9. (Original post by charco)
OK, so the equattion tells you that the moles of carbon dioxide will be equal to the moles of the limiting reagent.

MgCO3 is 0.00356 mol
H2SO4 is 0.0005 mol

which wuold make magnesium carbonate in excess ...

But 0.1 x 0.05 moles of sulphuric acid = 0.005 mol NOT 0.0005 mol
0.3 g of magnesium carbonate = 0.3/84 = 0.00357 mol

So, yes sulphuric acid is in excess and magnesium carbonate is the limiting reagent.

Hence moles of CO2 = moles of magnesium carbonate = 0.00357 mol.

1 mol of gas at RTP = 24000 cm3

hence 0.00357 mol = 0.00357 * 24000 = 85.7 cm3
Ah ok, thanks! Is it because of the 1:1 ratio that co2 = MgCO3

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10. (Original post by kath_igidbashian)
Ah ok, thanks! Is it because of the 1:1 ratio that co2 = MgCO3

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Yes, the carbon dioxide should be equal to the moles of the limiting reagent ...
11. Thank you so much for your help! You're a life saver!!

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