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. 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.

  2. 567.99

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