Answer:
11.2 grams CaO
Explanation:
It appears that substance X may be calcium carbonate: 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. 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. 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. Any less, then blame it on your lab partner. But don't try taking credit if you have more than 11.2 grams. Scraping debri off the counter into the beaker doesn't count.
