Question

what volume of carbon dioxide is produced when 20.5 g of calcium carbonate reacts completely according to the following reaction at 25°c and 1 atm?
calcium carbonate (s) → calcium oxide (s) + carbon dioxide (g)
volume = □ l
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Explanation:

Step1: Calculate the molar mass of calcium carbonate

The molar mass of \(CaCO_3\) is \(M = 40.08+12.01 + 3\times16.00=100.09\space g/mol\).
The number of moles of \(CaCO_3\), \(n=\frac{m}{M}=\frac{20.5\space g}{100.09\space g/mol}\approx0.205\space mol\)

Step2: Use the stoichiometry of the reaction

From the reaction \(CaCO_3(s)\to CaO(s)+CO_2(g)\), the mole ratio of \(CaCO_3\) to \(CO_2\) is \(1:1\). So, \(n_{CO_2}=n_{CaCO_3} = 0.205\space mol\)

Step3: Use the ideal gas law \(PV = nRT\)

Given \(P = 1\space atm\), \(T=25^{\circ}C=(25 + 273.15)K=298.15K\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\)
\(V=\frac{nRT}{P}=\frac{0.205\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times298.15K}{1\space atm}\)
\(V=\frac{0.205\times0.0821\times298.15}{1}\space L\approx5.01\space L\)

Answer:

\(5.01\space L\)