Answer:
11.2 grams CaO
Explanation:
It appears that substance X may be calcium carbonate: Â CaCO3
CaCO3 can be thermally decomposed to CO2 and CaO in the following balanced reaction:
CaCO3(s) ⇒ CaO(s) + CO2(g)   (with applied heat, 840°C)
The molar ratio between the product, CaO, and the reactant, CaCO3, is 1:1. Â If we start with 1 mole CaCO3, we should produce 1 mole of CaO.
We have 20.0 grams of substance X, which we'll label CaCO3. Â Calculate the moles of CaCO3 by using its molar mass of 100.1 grams/mole.
 20.0 grams/(100.1 grams/mole) = 0.1998 or 0.200 moles of CaCO3.
This should produce, with a molar ratio of 1 to 1, 0.200 moles of CaO
Convert this to grams CaO by multiply by it's molar mass of 56.1 g/mole:
(0.200 moles)*(56.1 g/mole) = 11.2 grams CaO. Â Any less, then blame it on your lab partner. Â But don't try taking credit if you have more than 11.2 grams. Â Scraping debri off the counter into the beaker doesn't count.