Ta có: \(n_{Al_4C_3}=\dfrac{72}{144}=0,5\left(mol\right)\)
Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{Al\left(lý.thuyết\right)}=4n_{Al_4C_3}=2\left(mol\right)\\n_{C\left(lý.thuyết\right)}=3n_{Al_4C_3}=1,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al\left(thực\right)}=\dfrac{2\cdot27}{60\%}=90\left(g\right)=b\\m_{C\left(thực\right)}=\dfrac{12\cdot1,5}{60\%}=30\left(g\right)=a\end{matrix}\right.\)