\(a/CTHH:Al_x\left(SO_4\right)_y\\ x.II=y.III\\ \Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\\ x=2;y=3\\ \Rightarrow CTHH:Al_2\left(SO_4\right)_3\\ b/2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ c/m=n.M\\ BTKL:m_{H_2SO_4}=1,5+85,5-13,5=73,5g\)
