a) \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
PTHH: \(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)
0,25--->\(\dfrac{0,25}{3}\)------>\(\dfrac{0,25}{3}\)
=> \(m_{FeCl_3}=\dfrac{0,25}{3}.162,5=\dfrac{325}{24}\left(g\right)\)
b) \(m_{Fe\left(OH\right)_3}=\dfrac{0,25}{3}.107=\dfrac{107}{12}\left(g\right)\)