Answer:
[tex]\boxed{\text {0.215 mol}}[/tex]
Explanation:
1 mol of CaCO₃ = 100.09 g
[tex]\text{Moles of CaCO$_{3}$ } = \text{21.5 g CaCO$_{3}$} \times \dfrac{\text{1 mol CaCO$_{3}$}}{\text{100.09 g CaCO$_{3}$}} = \textbf{0.215 mol CaCO$_{3}$}\\\text{There are $\boxed{\textbf{0.215 mol}}$ of CaCO$_{3}$ in 21.5 g}[/tex]
