Answer:
[tex]\boxed{\text{0.672 mol}}[/tex]
Explanation:
[tex]\text{M$_{r}$ of Al(C$_{3}$H$_{7}$)$_{3}$} = 156.24\\\text{Moles} = \text{105 g } \times \dfrac{\text{1 mol}}{\text{156.24 g}} = \text{0.672 mol}\\\\\rm \text{The sample contains $\boxed{\textbf{0.672 mol}}$ of Al(C$_{3}$H$_{7}$)$_{3}$}[/tex]
