Đáp án + Giải thích các bước giải:

Sửa: $3,16g$ hh A

$PTHH:$

$Mg+CuCl_2\to MgCl_2+Cu$

$Fe+CuCl_2\to FeCl_2+Cu$

$FeCl_2+2NaOH\to Fe(OH)_2\downarrow+2NaCl$

$MgCl_2+2NaOH\to Mg(OH)_2\downarrow+2NaCl$

$Mg(OH)_2\xrightarrow{t^o}MgO+H_2O$

$4Fe(OH)_2+O_2\xrightarrow{t^o}2Fe_2O_3+4H_2O$

Vì $m_A=3,16<m_{oxit}\Rightarrow $ Fe dư

Đặt $n_{Mg}=x(mol);n_{Fe(p/ứ)}=y(mol);n_{Fe(dư)}=z(mol)$

$\Rightarrow 24x+56y+56z=31,6(1)$

Theo PT: $n_{MgO}=n_{Mg}=x(mol);n_{Fe_2O_3}=0,5n_{Fe}=0,5y$

$\Rightarrow 40x+160.0,5y=1,4(2)$

Mặt khác: $m_C=m_{Cu}+m_{Fe(dư)}=64x+64y+56z=3,84(3)$

Từ $(1)(2)(3)\Rightarrow x=0,005(mol);y=0,015(mol)$

$\Rightarrow \%m_{Mg}=\dfrac{0,005.24}{3,16}.100\%=3,8\%$

$\Rightarrow \%m_{Fe}=100-3,8=96,2\%$

Theo PT: $n_{CuCl_2}=x+y=0,02(mol)$

$\Rightarrow C_{M_{CuCl_2}}=\dfrac{0,02}{0,25}=0,08M$