nAl=\(\frac{2,7}{27}=0,1\left(mol\right)\)
nCuO=\(\frac{8}{80}=0,1\left(mol\right)\)
a)2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 (1)
CuO + H2 -to-> Cu + H2O (2)
b) theo pt (1) : nH2=\(\frac{3}{2}.nAl=\frac{3}{2}.0,1=0,15\left(mol\right)\)
CuO + H2 -to-> Cu + H2O (2)
0,1 < 0,15
-> CuO hết, H2 dư
theo pt (2): nCu=nCuO=0,1 (mol)
-> mCu=0,1.64=6,4 (g)