$Fe$ bị thụ đồng trong $H_2SO_4$ đặc nguội

$n_{SO_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$\xrightarrow{BTe}n_{Cu}=n_{SO_2}=0,1(mol)$

$m_{Fe}=12-0,1×64=5,6(g)$

$n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$

$→\%Fe=\dfrac{5,6}{12}×100\%=46,66\%$

$→\%Cu=100-46,66=53,34\%$

$\xrightarrow{BTNT\ S}n_{BaSO_4}=n_{SO_2}=0,1(mol)$

$→m=0,1×233=23,3(g)$

Đáp án:

\(\%m_{Cu}≈53,33\%;\%m_{Fe}≈46,67\%\)

\(m=23,3\) 

Giải thích các bước giải:

\(\begin{array} {l} n_{SO_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Cu+2H_2SO_4\to CuSO_4+SO_2↑+2H_2O\\ SO_2+Br_2+2H_2O\to 2HBr+H_2SO_4\\ H_2SO_4+Ba(OH)_2\to BaSO_4↓+2H_2O\\ \text{Theo PT: }n_{SO_2}=n_{Cu}=0,1(mol)\\ \to \%m_{Cu}=\dfrac{0,1.64}{12}.100\%≈53,33\%\\ \to \%m_{Fe}=100-53,33=46,67\%\\ \text{Theo PT: }n_{BaSO_4}=n_{H_2SO_4}=n_{SO_2}=0,1(mol)\\ \to m=m_{BaSO_4}=0,1.233=23,3(g) \end{array}\)