Đáp án:

`%`$\rm m_{Cu}$`~~19,29%`

`%`$\rm m_{Zn}$`~~46,43%`

`%`$\rm m_{Mg}$`=34,28%`

Giải thích các bước giải:

$\rm a)$

              $\rm Zn+H_2SO_4\to ZnSO_4+H_2$

              $\rm Mg+H_2SO_4\to MgSO_4+H_2$

$\rm b)$

$\rm m_{cr}=m_{Cu}=2,7(g)$

$\rm Gọi\ n_{Zn}:a(mol)\ , n_{Mg}:b(mol)$

`=>`$\rm 65a+24b=14-2,7\ (1)$

$\rm n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

`=>`$\rm a+b=0,3\ (2)$

`(1)(2)=>`$\rm a=0,1(mol)\ , b=0,2(mol)$

`%`$\rm m_{Cu}$`={2,7}/{14}.100%~~19,29%`

`%`$\rm m_{Zn}$`={0,1.65}/{14}.100%~~46,43%`

`%`$\rm m_{Mg}$`=100%-19,29%-46,43%=34,28%`

Đáp án:

$\%m_{Ag}=19,29\%\\ \%m_{Zn}=46,43\%\\ \%m_{Mg}=34,28\%$

Giải thích các bước giải:

$a,\\ Zn+H_2SO_4\to ZnSO_4+H_2\\ Mg+H_2SO_4\to MgSO_4+H_2\\ b,\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ m_{Ag}=2,7(g)\\ Đặt\ \begin{cases}n_{Zn}=x(mol)\\n_{Mg}=y(mol)\end{cases}\\ \Rightarrow 65x+24y=14-2,7=11,3\ (1)\\ \xrightarrow{\text{Từ PTH}}\ n_{H_2}=n_{Zn}+n_{Mg}=x+y(mol)\\ \Rightarrow x+y=0,3\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,1\\y=0,2\end{cases}\\ \%m_{Ag}=\dfrac{2,7}{14}.100\%=19,29\%\\ \%m_{Zn}=\dfrac{0,1.65}{14}.100\%=46,43\%\\ \%m_{Mg}=\dfrac{0,2.24}{14}.100\%=34,28\%$