The reaction for the burning of calcium carbonate is:
\[
\text{CaCO}_3 \xrightarrow{\text{heat}} \text{CaO} + \text{CO}_2
\]
From the reaction, we know the molar mass of calcium carbonate (CaCO\(_3\)) is 100 g/mol, and calcium oxide (CaO) is 56 g/mol.
The stoichiometric ratio is 1:1, meaning that 100 g of calcium carbonate gives 56 g of calcium oxide.
Therefore, if 28 g of calcium oxide is produced, the mass of calcium carbonate required is:
\[
\frac{100}{56} \times 28 = 50 \, \text{g}
\]
Thus, the correct answer is (d).