Bạn tham khảo!

$\text{Gọi kim loại hóa trị III là A}$

$\text{2A+6HCl$\rightarrow$2ACl3+3H2}$

$\text{nH2=$\frac{V}{22,4}$=$\frac{2,688}{22,4}$=0,12 mol}$

$\text{Có nA=$\frac{0,12.2}{3}$=0,08 mol}$

$\rightarrow$ $\text{M A=$\frac{m}{n}$=$\frac{2,16}{0,08}$=27}$

$\text{a) Vậy KL A là Nhôm (Al)}$

$\text{Có nHCl=$\frac{0,12.6}{3}$=0,24 mol}$

$\rightarrow$ $\text{mHCl=n.M=0,24.36,5=8,76g}$

$\rightarrow$ $\text{b) C% HCl=$\frac{\text{mct}}{\text{mdd}}$=$\frac{8,76}{120}$.100=7,3%}$

$\text{Có nAlCl3=nAl=0,08 mol}$

$\rightarrow$ $\text{mAlCl3=n.M=0,08.133,5=10,68g}$

$\text{c) C% AlCl3=$\frac{\text{mct}}{\text{mdd}}$=$\frac{10,68}{120}$.100=8,9%}$

#TRANPHAMGIABAO

Bạn tham khảo:

$a/$

$2R+6HCl \to 2RCl_3+3H_2$

$n_{H_2}=\dfrac{2,688}{22,4}=0,12(mol)$

$n_R=\dfrac{2}{3}.n_{H_2}=0,08(mol)$

$M_R=\dfrac{2,16}{0,08}=27 (Al)$

$b/$

$n_{HCl}=0,12.2=0,24(mol)$

$C\%_{HCl}=\dfrac{0,24.36,5}{120}.100\%=7,3\%$

$c/$

$n_{AlCl_3}=0,08(mol)$

$C\%_{AlCl_3}=\dfrac{0,08.133,5}{2,16+120-0,12.2}.100\%=8,8\%$