a) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,2<--0,4<-----------0,2
=> mMg = 0,2.24 = 4,8 (g)
=> \(\%m_{Mg}==\dfrac{4,8}{12,8}.100\%=37,5\%\)
=> \(\%m_{CuO}=100\%-37,5\%=62,5\%\)
b) \(n_{CuO}=\dfrac{12,8.62,5\%}{80}=0,1\left(mol\right)\)
PTHH: CuO + 2HCl --> CuCl2 + H2O
0,1-->0,2
=> nHCl = 0,2 + 0,4 = 0,6 (mol)
=> mHCl = 0,6.36,5 = 21,9 (g)
=> \(C\%_{dd.HCl}=\dfrac{21,9}{200}.100\%=10,95\%\)