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group 2 mole calculations help!!

1.39g of magnesium carbonate are added to 100cm3 of sulfuric acid.
MgCO3 (s) + H2SO4 (aq) -> MgSO4 (aq) + CO2 (g) + H2O (l)
During the reaction, 212cm3 of gas were collected. Determine which reactant was in excess. Show your working.

i worked out the moles of MgCO3 to be 0.0165. how do i work out the moles of H2SO4 ? i assume it has something to do with the gas collected and the ratios but i dont know what to do

and also this question please!!


Student 1 heated a sample of an unknown group 2 metal strongly in a crucible. after heating, a metal oxide was formed. the crucible and the metal were weighed before, and after heating. the results are shown as below
mass of empty crucible and lid (g) = 8.896
mass of crucible and lid with metal (g) = 10.256
mass of crucible and lid with metal oxide after heating (g) =10.817

a. write balanced chemical equation, including state symbols, for the reaction. use M to represent the unknown metal.
O2 (g) + M (s) -> MO (s)

b. calculate the moles of oxygen that reacted with the metal.
0.0350

c. how many moles of metal were present?

d. determine the molar mass of the metal and identify the metal.

how would i work out how many moles of the unknown metal were present??
Reply 1
1 mol of gas occupies 24dm3. Use this to try and work out moles of CO2 :smile:

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Reply 2
Original post by Dylann
1 mol of gas occupies 24dm3. Use this to try and work out moles of CO2 :smile:

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ahhh i might be being really stupid but i still cant work out what to do .....
Reply 3
Original post by Kyosu
ahhh i might be being really stupid but i still cant work out what to do .....


Not at all, it's a difficult question.

If 1 mol of gas occupies 24dm3:

1mol=24'000cm3
xmol=212cm3

Find x

Or you could use, 1cm3 = 1/24'000mol
x=1/24'000mol x 212.
= 212/24'000 mol of CO2

Then use ratios


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