When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction
\rm CaCO_3{\it(s)}\rightarrow CaO{\it(s)}+CO_2{\it(g)}
How many grams of calcium carbonate are needed to produce 53.0 L of carbon dioxide at STP?
2 answers
-
1. CaCO3(s) ==> CaO(s) + CO2(g)
2. Convert 53.0 L to moles. moles = 53.0/22.4.
3. Using the coefficients in the balanced equation, convert moles CO2 to moles CaCO3.
4. Now convert moles CaCO3 to grams. g = moles x molar mass.
-
567.99
