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danhirons
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Report Thread starter 9 years ago
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Procedure
A 50.0 cm3 sample of 0.100 mol dm–3 sulfuric acid was measured using a pipette. This was
added 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 a
trough of water.
The two pieces of apparatus were connected by a delivery tube so that any gas produced in the
side arm flask would be collected in the measuring cylinder. The apparatus was airtight and no
leaks were present.
The rubber bung was removed from the side arm flask. The MgCO3 was quickly added and the
bung replaced.
Bubbles of carbon dioxide were seen collecting in the measuring cylinder.
When no more gas was seen to collect in the measuring cylinder, the volume of gas collected
was recorded as 64 cm3.

Calculate the number of moles of H2SO4 and MgCO3 used in the experiment.




For some reason the answer to MgCO3 is 0.00298 mol which I can't seem to get myself.
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charco
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Report 9 years ago
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Using the volume of gas made...

2H+ + CO32- --> H2O + CO2

assume 1 mole = 24 dm3 at RTP

60cm3 = 64/24000 = 0.00267 mol

therefore moles of carbonate used = moles of sulphuric acid used = 0.00267 moles

now assuming that the conditions are STP

64cm3 = 64/22414 = 0.00286 mol

moles of acid added = 0.1 x 0.05 = 0.005
moles of MgCO3 added = 0.3/84 = 0.00357

Magnesium carbonate is the limiting reagent, so will produce the same number of moles of carbon dioxide = 0.00357...

Unless you are omitting some pressure/temperature information the question is wrong!
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danhirons
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Report Thread starter 9 years ago
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So you get the number of moles of MgCO3 as 0.00357?

That's what I got and thought the question must be wrong. It's from an official OCR specimen paper as well!
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