The answer is CaSO3 -> CaO + SO2. I understand that when SO3 is in a vacuum...it turns to SO2...but I don't get how you predict the other product (CaO)...is it just a format that I just have to memorize?
2006-12-28 17:58:31 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
What if it wasn't in a vacuum?
2006-12-28 18:12:56 · update #1
If CaCO3 is heated in a vacuum, then no other material can enter into the reaction. That makes this a DECOMPOSITION reaction. In a decomposition reaction you have one compound in the reactants and two (generally) in the products.
Since you already know that the sufite will decompose into sulfur dioxide, you simply put the remaining materials together and balance by charges.
CaSO3 ---> SO2 (g) + CaO (Ca 2+ and O 2- don't require subscripts.)
2006-12-28 18:10:38 · answer #1 · answered by physandchemteach 7 · 0⤊ 0⤋
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Predict the reaction: solid calcium sulfite is heated in a vacuum?
The answer is CaSO3 -> CaO + SO2. I understand that when SO3 is in a vacuum...it turns to SO2...but I don't get how you predict the other product (CaO)...is it just a format that I just have to memorize?
2015-08-06 06:07:44 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Calcium Carbonate ---(heat)---> Calcium Oxide + Carbon Dioxide This reaction is called Thermal Decomposition
2016-03-18 14:32:02 · answer #3 · answered by ? 3 · 0⤊ 0⤋
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it is a thermal decomposition reaction and part of the limestone cycle. when calcium carbonate is thermally decomposed, it forms calcium oxide and the waste gas that forms is carbon dioxide: CaCO3 ==(heat)==> CaO + CO2 calcium carbonate ==(heat)==> calcium oxide + carbon dioxide
2016-04-07 04:23:14 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Calcium sulfite is a salt of calcium oxide (hydroxide) and sulfurous acid (H2SO3 or SO2 if dehydrated). Think of it that way. You can't split CaO into calcium metal except at super high temperatures.
SO3 (sulfur trioxide) will form if calcium sulFATE is heated (CaSO4)
2006-12-28 18:07:15 · answer #5 · answered by Anonymous · 0⤊ 0⤋