$D_{Fe}=7,8(g/cm^3)$
$\to V_{Fe_{\text{bđ}}}=\dfrac{2,8}{7,8}=\dfrac{14}{39}(cm^3)$
$\to R_{Fe\text{bđ}}=0,44(cm)$
$n_{Fe}=\dfrac{2,8}{56}=0,05(mol)$
$n_{HCl}=0,5.0,175=0,0875(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$\to n_{Fe\text{dư}}=6,25.10^{-3}(mol)$
$\to m_{Fe\text{dư}}=0,35g$
$\to V_{Fe\text{dư}}=\dfrac{7}{156}(cm^3)$
$\to R_{Fe\text{dư}}=0,22(cm)$
Vậy bán kính viên bi còn lại bằng $\dfrac{0,22}{0,44}=\dfrac{1}{2}$ ban đầu.
