Đáp án + Giải thích các bước giải:

$a)$ Đặt $n_{Fe}=n_{Mg}=x(mol)$

$\Rightarrow n_{Zn}=\dfrac{13-56x-24x}{65}=\dfrac{13-80x}{65}(mol)$

Mặt khác: $x+x+\dfrac{13-80x}{65}=0,3$

$\Rightarrow x=0,13(mol)$

$\Rightarrow n_{Fe}=n_{Mg}=0,13(mol);n_{Zn}=0,3-0,13.2=0,04(mol)$

$\Rightarrow \begin{cases} m_{Fe}=0,13.56=7,28(g)\\ m_{Mg}=0,13.24=3,12(g)\\ m_{Zn}=0,04.65=2,6(g) \end{cases}$

$\Rightarrow \begin{cases} \%m_{Fe}=\dfrac{7,28}{13}.100\%=56\%\\ \%m_{Mg}=\dfrac{3,12}{13}.100\%=24\%\\ \%m_{Zn}=100-56-24=20\% \end{cases}$

$b)CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Bảo toàn electron: $n_{H_2}=n_{Mg}+n_{Fe}+n_{Zn}=0,3(mol)$

Xét tỉ lệ: $n_{CuO}=\dfrac{80}{80}=1>n_{H_2}\Rightarrow CuO$ dư

Theo PT: $n_{Cu}=n_{H_2}=n_{CuO(pứ)}=0,3(mol)$

$\Rightarrow m_{\rm CR}=0,3.64+(1-0,3).80=75,2(g)$

Đáp án:

$\rm \%m_{Fe}=56\%$

$\rm \%m_{Mg}=24\%$

$\rm \%m_{Zn}=20\%$

$\rm m_{Chất\ rắn}=75,2\ (g)$

Giải thích các bước giải:

Đặt $\rm n_{Zn}=a\ (mol),$ $\rm n_{Fe}=n_{Mg}=b\ (mol)$

Phương trinh khối lượng:

$\rm 65a+(56+24)b=13\ (g)\ (1)$

$\rm n_{A}=a+2b=0,3\ (mol)\ (2)$

Từ $\rm (1)(2)$ giải được:

$\begin{cases} \rm a=0,04\ (mol)\\\rm b=0,13\ (mol) \end{cases}$

$\rm \to \%m_{Fe}=\dfrac{0,13.56}{13}.100\%=56\%$

$\rm \to \%m_{Mg}=\dfrac{0,13.24}{13}.100\%=24\%$

$\rm \to \%m_{Zn}=100\%-56\%-24\%=20\%$

$\rm PTHH:$

$\rm Fe+2HCl \to FeCl_2+H_2$

$\rm Mg+2HCl \to MgCl_2+H_2$

$\rm Zn+2HCl \to ZnCl_2+H_2$

Theo PTHH:

$\rm n_{H_2}=n_{Fe}+n_{Mg}+n_{Zn}=0,13+0,13+0,04=0,3\ (mol)$

$\rm n_{CuO}=\dfrac{80}{80}=1\ (mol)$

$\rm PTHH:$

$\rm H_2+CuO \xrightarrow{t^o}Cu+H_2O$

Tỉ lệ: $\rm \dfrac{0,3}{1}<\dfrac{1}{1}$

Suy ra $\rm n_{CuO}$ dư, tính theo $\rm n_{H_2}.$

$\rm m_{Cu}=64.0,3=19,2\ (g)$

$\rm m_{CuO\ dư}=80.(1-0,3)=56\ (g)$

$\rm \to m_{Chất\ rắn}=19,2+56=75,2\ (g)$