\(nMg=\dfrac{3,6}{24}=0,15\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
1 2 1 1 (mol)
0,15 0,3 0,15 0,15 (mol)
\(mHCl=0,3.36,5=10,95\left(g\right)\)
=> HCl đủ , Mg đủ
\(H_2+CuO\rightarrow Cu+H_2O\)
1 1 1 1 (mol)
0,15 0,15 (mol)
=> \(mCu=0,15.64=9,6\left(g\right)\)