Đáp án:

 m=3,6 gam

mAlCl3=16,02 gam

C% AlCl3=13,02%

Giải thích các bước giải:

\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)

Ta có: \({m_{HCl}} = 14,6\% .120 = 17,52{\text{ gam}} \to {{\text{m}}_{HCl}} = \frac{{17,52}}{{36,5}} = 0,48{\text{ mol}}\)

\( \to {n_{Al}} = {n_{AlC{l_3}}} = \frac{1}{3}{n_{HCl}} = 0,12{\text{ mol}}\)

\( \to {m_{Al}} = 0,12.27 = 3,24{\text{ gam}} \to {\text{m = }}\frac{{{m_{Al}}}}{{90\% }} = 3,6{\text{ gam}}\)

\( \to {{\text{m}}_{AlC{l_3}}} = 0,12.(27 + 35,5.3) = 16,02{\text{ gam}})

BTKL: \({m_{dd{\text{ A}}}} = m + {m_{ddHCl}} - {m_{{H_2}}} = 3,6 + 120 - 0,3.2 = 123{\text{ gam}} \to {\text{C}}{{\text{\% }}_{AlC{l_3}}} = \frac{{16,02}}{{123}} = 13,02\% \)

a.

2Al+6HCL->2AlCl3+3H2

b.

Ta có lượng HCl thực tế: 14,6*120/100=17,52g HCl

nHCl=17.52/27=0.65mol; 

nAl=1/3nHCl=0.65/3=0.2mol

nAlCl2=nAl=0.2mol

=>mAl=0.2*27=5.4gam

mAl nguyên chất: 5.4-5.4*10/100=4.86gam

=>mAlCl3=0.2*133.5=26.7gam

c.