\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\
pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,3 0,45
\(C_{M\left(H_2SO_4\right)}=\dfrac{0,45}{0,3}=1,5M\\
n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(LTL:\dfrac{0,5}{1}>\dfrac{0,45}{1}\)
=> CuO dư
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,45\left(mol\right)\\
m_{Cr}=\left(0,5-0,45\right).80+0,45.64=32,8g\)