\(a,m_{ddCuSO_4}=\dfrac{24.100}{15}=160\left(g\right)\)
\(b,m_{\left(Al\left(NO_3\right)_3\right)}=0,5.89=44,5\left(g\right)\)
\(m_{ddAl\left(NO_3\right)}=\dfrac{44,5.100}{4}=1112,5\left(g\right)\)
\(c,n_{FeCl_3}=0,8.0,15=0,12\left(mol\right)\)
\(m_{FeCl_3}=0,12.106,5=12,78\left(g\right)\)