Đáp án:

$a)PTHH:Mg+2HCl\to MgCl_2+H_2\uparrow$

$2Al+6HCl\to 2AlCl_3+3H_2\uparrow$

$2Cu+O_2\xrightarrow{t^o}2CuO$

$CuO+2HCl\to CuCl_2+H_2O$

$\%Mg=20,81$

$\%Al=5,2$

$\%Cu=73,99$

$b)C_{M_X}=\dfrac{4}{7}M$

$C_{M_Y}=\dfrac{8}{7}M$

Giải thích các bước giải:

$a)PTHH:Mg+2HCl\to MgCl_2+H_2\uparrow$

$2Al+6HCl\to 2AlCl_3+3H_2\uparrow$

$2Cu+O_2\xrightarrow{t^o}2CuO$

$CuO+2HCl\to CuCl_2+H_2O$

Cần chung lượng axit $\to$ sinh ra cùng lượng $H_2$

$n_{H_2}=\dfrac{0,3584}{22,4}=0,016(mol)$

$\Rightarrow n_{Cu}=n_{H_2}=0,016(mol)$

$\Rightarrow m_{Cu}=0,016.64=1,024(g)$

Đặt $n_{Mg}=x(mol);n_{Al}=y(mol)$

$\Rightarrow x+1,5y=0,016$

Mặt khác: $24x+27y=1,384-1,024=0,36$

$\Rightarrow x=0,012(mol);y\approx 0,00027(mol)$

$\Rightarrow m_{Mg}=0,012.24=0,288(g)$

$\Rightarrow \%m_{Mg}=\dfrac{0,288}{1,384}.100\%\approx 20,81\%$

$\Rightarrow \%m_{Cu}=\dfrac{1,024}{1,384}.100\%=73,99\%$

$\Rightarrow \%m_{Al}=100-73,99-20,81=5,2\%$

$b)$ Đặt $n_{HCl(X)}=a(mol);n_{HCl(Y)}=b(mol)$

Đặt $C_{M_X}=c(M)\Rightarrow C_{M_Y}=2c(M)$

Ta có $\dfrac{1}{2}n_{Al}=\dfrac{1}{750}(mol)$

$\Rightarrow a=3.6.\dfrac{1}{2}n_{Al}=0,024(mol)$

Mặt khác: $n_{HCl}=2n_{H_2}=0,032(mol)$

$\Rightarrow b=0,032-0,024=0,08(mol)$

$\Rightarrow V_{X}=\dfrac{0,024}{c};V_Y=\dfrac{0,008}{2c}=\dfrac{0,004}{c}$

$\Rightarrow \dfrac{0,024}{c}+\dfrac{0,008}{c}=0,056$

$\Rightarrow c=\dfrac{4}{7}$

Vậy $C_{M_X}=\dfrac{4}{7}M$ và $C_{M_Y}=\dfrac{8}{7}M$