\(a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\b. n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\\ \Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\\ \Rightarrow m_{Cu}=10-5,4=4,6\left(g\right)\\ c.n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\\ \Rightarrow m_{ddH_2SO_4}=\dfrac{0,3.98}{20\%}=147\left(g\right)\)