$1atm=1,01325bar$
$\to 1bar=\dfrac{1}{1,01325}atm$
$n_{HCl}=\dfrac{100,85.1,19.36,5\%}{36,5}=1,2(mol)$
$n_{SO_2}=\dfrac{\dfrac{1}{1,01325}.11,2}{0,082.(25+273)}=0,45(mol)$
$n_{Cu(OH)_2}=\dfrac{39,2}{98}=0,4(mol)$
$CuO+2HCl\to CuCl_2+H_2O$
$Cu+2H_2SO_4\to CuSO_4+SO_2+2H_2O$
$HCl+NaOH\to NaCl+H_2O$
$2NaOH+CuCl_2\to Cu(OH)_2+2H_2O$
$n_{Cu}=n_{SO_2}=0,45(mol)$
$n_{Cu(OH)_2}=n_{CuCl_2}=n_{CuO\text{pứ}}=0,4(mol)$
Ta có $0,4.2=0,8<1,2\to HCl$ dư
$\to CuO$ tan hết
Vậy:
$\%m_{Cu}=\dfrac{0,45.64.100}{0,45.64+0,4.80}=47,37\%$
$\%m_{CuO}=52,63\%$