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Calculation of expected gas volume. Please help!

Hi guys! I trying to do Evaluative Task OCR past paper (2008-2009) and really stuck with the expected gas volume calculation.
A 50.0 cm3 sample of 0.100 mol dm–3 sulfuric acid was measured using a pipette. This wasadded to the flask with the side arm.
0.30 g of magnesium carbonate, MgCO3, was accurately weighed on a digital balance.
The measuring cylinder was filled with exactly 100 cm3 of water and supported upside down in atrough of water.

When no more gas was seen to collect in the measuring cylinder, the volume of gas collectedwas recorded as 64 cm3.

Show by calculation that the expected volume of carbon dioxide is greater than the 64 cm3
obtained.
Assume that one mole of CO2 occupies 24.0 dm3 under the conditions used in theexperiment.







H2SO4(aq) + MgCO3(s) → MgSO4(aq) + H2O(l) + CO2(g)

Step 1. I found number of moles of 0.30/84.3=3.56x10-3 which is right
Step 2. Ratio from the formula 1 mole of MgCO3 = 1mole of CO2
so 3.56x10-3 x 24000=85.44 cm2 and in the mark scheme the answer is 71.4 cm3. I don't understand why!!!
I would really appreciate if someone could point me in the right direction.
Original post by Prada:)
Hi guys! I trying to do Evaluative Task OCR past paper (2008-2009) and really stuck with the expected gas volume calculation.
A 50.0 cm3 sample of 0.100 mol dm–3 sulfuric acid was measured using a pipette. This wasadded to the flask with the side arm.
0.30 g of magnesium carbonate, MgCO3, was accurately weighed on a digital balance.
The measuring cylinder was filled with exactly 100 cm3 of water and supported upside down in atrough of water.

When no more gas was seen to collect in the measuring cylinder, the volume of gas collectedwas recorded as 64 cm3.

Show by calculation that the expected volume of carbon dioxide is greater than the 64 cm3
obtained.
Assume that one mole of CO2 occupies 24.0 dm3 under the conditions used in theexperiment.







H2SO4(aq) + MgCO3(s) → MgSO4(aq) + H2O(l) + CO2(g)

Step 1. I found number of moles of 0.30/84.3=3.56x10-3 which is right
Step 2. Ratio from the formula 1 mole of MgCO3 = 1mole of CO2
so 3.56x10-3 x 24000=85.44 cm2 and in the mark scheme the answer is 71.4 cm3. I don't understand why!!!
I would really appreciate if someone could point me in the right direction.


You are correct the mark scheme is wrong, our teacher said so.
Reply 2
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Prada:)
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7
Original post by Assassin2015
You are correct the mark scheme is wrong, our teacher said so.

Thanks a lot! That makes sense now )))
Reply 3
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desreanna
jdfsg

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