Bạn tham khảo:

$a/$

P1:

$Fe+H_2SO_4 \to FeSO_4+H_2$

$n_{H_2}=0,15(mol)$

$\to n_{Fe}=0,15(mol)$

$\to n_{Fe(bđ)}=0,15.3=0,45(mol)$

P2:

$n_{SO_2}=0,3(mol)$

Bt $e$:

$3.n_{Fe}+2.n_{Cu}=2.n_{SO_2}$

$3.0,15+2.n_{Cu}=2.0,3$

$\to n_{Cu}=0,075(mol)$
$\to n_{Cu(bd)}=0,075.3=0,225(mol)$

$a=m_X=0,225.64+0,45.56=19,6(g)$

P3:

$Fe+2AgNO_3 \to Fe(NO_3)_2+2Ag$

$\to n_{Ag}=0,3(mol)$

$Cu+2AgNO_3 \to Cu(NO_3)_2+2Ag$

$\to n_{Ag}=0,15(mol)$

$n_{AgNO_3}=0,5(mol)$

BT $e$:

$2.n_{Fe}+2.n_{Cu}=n_{Ag}$

$2.0,15+2.0,075=0,5$

$\to 0,45=0,5(loại)$

$\to$ sau phản ứng có $AgNO_3$ dư

$Fe(NO_3)_2+AgNO_3 \to Fe(NO_3)_3+Ag$

$n_{AgNO_3}=0,5-0,45=0,05(mol)$

$\to n_{Ag}=0,05+0,15+0,3=0,5(mol)$

$m_{Ag}=0,5.108=54(g)$

$b/$
$Cu+2Fe(NO_3)_3 \to Cu(NO_3)_2+2Fe(NO_3)_2$

$n_{Cu}=0,025(mol)$

$m_{Cu}=0,025.64=1,6(g)$

Đáp án:

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