\(a,n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ n_{CuSO_4}=\dfrac{24}{160}=0,15(mol)\\ PTHH:2Al+3CuSO_4\to Al_2(SO_4)_3+3Cu\)
Vì \(\dfrac{n_{Al}}{2}>\dfrac{n_{CuSO_4}}{3}\) nên \(Al\) dư
\(n_{Al(dư)}=0,2-\dfrac{2}{3}.0,15=0,1(mol)\\ \Rightarrow m_{Al(dư)}=0,1.27=2,7(g)\\ b,n_{Cu}=n_{CuSO_4}=0,15(mol)\\ \Rightarrow m_{Cu}=0,15.64=9,6(g)\)
