Question

A sample of 30.695 g of limestone (calcium carbonate) is heated at high temperature and decomposes according to the calcination reaction
After a warm-up, the limestone suffered complete decomposition and after the cold was weighed and indicated 30.14 g . Determine the volume of gas released at STP
bobpursley
complete decomposition? I doubt it.
CaCO3>>CaO + CO2
so how many moles is (30.695-30.14)g?

moles=mass/44
lastly, volume at stp= moles*22.4 liters
I am not certain of the equation from your post. The warm up of what? The cold what was weighted? I will assume the reaction is
CaCO3(s) + heat ==> CaO + CO2
mass lost on heating is 30.695-30.14 = ? = mass CO2 produced.
mols CO2 = grams/molar mass = ? = n
You know 1 mol of gas @ STP has a volume of 22.4 L. Convert mols to L OR you may use the gas law, PV = nRT to solve for volume in liters.

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