Đáp án:

$C$

Giải thích các bước giải:

Câu 23:

Xét $10$ gam hh X:

$\Rightarrow m_{Cu}=0,64(g)$

$\Rightarrow m_{Cr}+m_{Al}+m_{Zn}=10-0,64=9,36(g)$

Ta có $m_{muối}=m_{KL}+m_{Cl^-}$

$\Rightarrow m_{Cl^-}=37,2-9,36=27,84(g)$

$\Rightarrow n_{Cl^-(1)}=\dfrac{27,84}{35,5}=0,784(mol)$

Mặt khác: $n_{Cl^-(1)}=2n_{Cr}+3n_{Al}+2n_{Zn}=0,784(1)$

Xét $20$ gam hh X:

$\Rightarrow m_{Cu}=0,64.2=1,28(g)$

$\Rightarrow n_{Cu}=\dfrac{1,28}{64}=0,02(mol)$

$\Rightarrow m_{Cl^-(2)}=m_{muối}-m_{KL}=79,94-20=59,94(g)$

$\Rightarrow n_{Cl^-(2)}=\dfrac{59,94}{35,5}=1,68845(mol)$

Có $n_{Cl^-(2)}=n_{\text{Cl trong muối Cr,Al,Zn}}+2n_{\text{Cl trong muối Cu}}$

$\Rightarrow n_{\text{Cl trong muối Cr,Al,Zn}}=1,68845-2.0,02=1,64845(mol)$

Trong 10 gam X có $n_{\text{Cl trong muối Cr,Al,Zn}}=\dfrac{1,64845}{2}=0,824(mol)$

$\Rightarrow 3n_{Cr}+3n_{Al}+2n_{Zn}=0,824(2)$

Lấy $(2)-(1)\Rightarrow n_{Cr}=0,04(mol)$

$\Rightarrow \%m_{Cu}=\dfrac{0,64}{10}.100\%=6,4\%$

$\Rightarrow \%m_{Cr}=\dfrac{0,04.52}{10}.100\%=20,8\%$