\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0,15
Cu không pứ với axit loãng.
\(\%m_{Al}=\dfrac{27.0,1.100}{9}=30\%\)
=> \(\%m_{Cu}=100-30=70\%\)
\(CM_{H_2SO_4}=\dfrac{0,15}{0,3}=0,5M\)