Câu 12:

$a,$

$n_{SO_2}=\dfrac{6,4}{64}=0,1(mol)$

$Cu+2H_2SO_4(đ)\xrightarrow{t^o}CuSO_4+SO_2+2H_2O$

$CuO+H_2SO_4(đ)\xrightarrow{t^o}CuSO_4+H_2O$

$b,$

Theo PT: $n_{Cu}=n_{SO_2}=0,1(mol)$

$→\%Cu=\dfrac{0,1×64}{14,4}×100\%=44,44\%$

$→\%CuO=100-44,44=55,56\%$

$c,$

$n_{NaOH}=0,1×1=0,1(mol)$

Đặt $T=\dfrac{n_{OH}}{n_{SO_2}}=1$ $\xrightarrow{}$ Tạo ra muối $NaHSO_3$

$\xrightarrow{BTNT\ S}n_{NaHSO_3}=0,1(mol)$

$→m_{NaHSO_3}=0,1×104=10,4(g)$

  Câu 13:

$a,$

$n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$

$n_{SO_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$2Al+6HCl\xrightarrow{}2AlCl_3+3H_2$

$Cu+2H_2SO_4(đ)\xrightarrow{}CuSO_4+SO_2+2H_2O$

$Cu$ không phản ứng với $HCl$, $Al$ không phản ứng với $H_2SO_4$ đặc nguội

Theo PT: $n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}×0,15=0,1(mol);\ n_{Cu}=n_{SO_2}=0,2(mol)$

$→m=m_{Al}+m_{Cu}=0,1×27+0,2×64=15,5(g)$