Asked by damon
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
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
Answers
Answered by
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
CaCO3>>CaO + CO2
so how many moles is (30.695-30.14)g?
moles=mass/44
lastly, volume at stp= moles*22.4 liters
Answered by
DrBob222
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.