Đáp án:

\(a,\ \%m_{Al}=28,42\%\\ \%m_{Al_2O_3}=71,58\%\\ b,\ m_{\text{dd HCl}}=262,5\ g.\\ c,\ C\%_{AlCl_3}=16,91\%\)

Giải thích các bước giải:

\(a,\ PTHH:\\ 2Al+6HCl\to 2AlCl_3+3H_2↑\ (1)\\ Al_2O_3+6HCl\to 2AlCl_3+3H_2O\ (2)\\ n_{H_2}=\dfrac{5,04}{22,4}=0,225\ mol.\\ Theo\ pt\ (1):\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,15\ mol.\\ ⇒m_{Al}=0,15\times 27=4,05\ g.\\ ⇒\%m_{Al}=\dfrac{4,05}{14,25}\times 100\%=28,42\%\\ ⇒m_{Al_2O_3}=14,25-4,05=10,2\ g.\\ ⇒\%m_{Al_2O_3}=\dfrac{10,2}{14,25}\times 100\%=71,58\%\\ b,\ n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\ mol.\\ Theo\ pt\ (1):\ n_{HCl}=2n_{H_2}=0,45\ mol.\\ Theo\ pt\ (2):\ n_{HCl}=6n_{Al_2O_3}=0,6\ mol.\\ ⇒m_{HCl}=(0,45+0,6)\times 36,5=38,325\ g.\\ ⇒m_{\text{dd HCl}}=\dfrac{38,325}{14,6\%}=262,5\ g.\\ c,\ ∑n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,15+0,2=0,35\ mol.\\ m_{\text{dd spư}}=m_{\text{hh}}+m_{\text{dd HCl}}-m_{H_2}=\\ \Rightarrow m_{\text{dd spư}}=14,25+262,5-0,225\times 2=276,3\ g.\\ ⇒C\%_{AlCl_3}=\dfrac{0,35\times 133,5}{276,3}\times 100\%=16,91\%\)

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Bài 4:

a.

n H2=$\frac{5,04}{22,4}$=0,225 mol

2Al + 6HCl → 2AlCl3 + 3H2↑

0,15   0,45       0,15    ←0,225         mol

Al2O3 + 6HCl → 2AlCl3 + 3H2O

 0,1→      0,6        0,2                       mol

m Al=0,15.27=4,05 g

m Al2O3=m hh-m Al

              =14,25-4,05 =10,2 g

⇒n Al2O3=$\frac{10,2}{102}$=0,1 mol

⇒%m Al=$\frac{4,05}{14,25}$.100≈28,42 %

   %m Al2O3=100%-28,42 %=71,58 %

b.

∑ n HCl=0,45+0,6=1,05 mol

m ct HCl=1,05.36,5=38,325 g

⇒mdd HCl=$\frac{38,325.100}{14,6}$=262,5 g

c.

-dd sau pứ:AlCl3

-mdd sau pứ=m hh+mdd HCl-m H2

                   =14,25+262,5-0,225.2 =276,3 g

∑n AlCl3=0,15+0,2=0,35 mol

m ct AlCl3=0,35.133,5=46,725 g

⇒C% AlCl3=$\frac{46,725}{276,3}$.100≈16,91 %

-----------------------Nguyễn Hoạt-----------------------