PTHH: Al2O3+6HCl➝2AlCl3+3H2O(1)

a)nAl2O3=\(\dfrac{10,2}{102}\)=0,1(mol)

   mHCl=\(\dfrac{5\%.219}{100\%}\)=10,95(g)

   ⇒nHCl=\(\dfrac{10,95}{36,5}\)=0,3(mol)

Xét tỉ lệ Al2O3:\(\dfrac{0,1}{1}\)=0,1

Xét tỉ lệ HCl:\(\dfrac{0,3}{6}\)=0,05

⇒HCl pứng hết,Al2O3 còn dư

Theo PTHH(1) ta có nAl2O3 pứng=\(\dfrac{nHCl}{6}\)=\(\dfrac{0,3}{6}\)=0,05(mol)

⇒nAl2O3 dư=nAl2O3ban đầu-nAl2O3 pứng=0,1-0,05=0,05(mol)

⇒mAl2O3 dư=0,05.102=5,1(g)

b) C%HCl=\(\dfrac{0,3.36,5}{219+10,2}\).100%=4,8%

     nAlCl3=0,1(mol)

⇒C%AlCl3=\(\dfrac{0,1.136,5}{10,2+219}\).100%=6%

Bài 1:

a: \(\sqrt{\left(\sqrt3-1\right)^2}\cdot\sqrt3=\sqrt3\left(\sqrt3-1\right)=3-\sqrt3\)

b: \(\sqrt{6+2\sqrt5}-\sqrt{6-2\sqrt5}\)

\(=\sqrt{\left(\sqrt5+1\right)^2}-\sqrt{\left(\sqrt5-1\right)^2}\)

\(=\sqrt5+1-\left(\sqrt5-1\right)=\sqrt5+1-\sqrt5+1=1+1=2\)

c: \(\sqrt{\left(2-\sqrt3\right)^2}+\sqrt{\left(1-\sqrt3\right)^2}\)

\(=2-\sqrt3+\left|1-\sqrt3\right|\)

\(=2-\sqrt3+\sqrt3-1=2-1=1\)

d: \(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)

\(=\sqrt{9^2-17}\)

\(=\sqrt{81-17}=\sqrt{64}=8\)

Bài 2:

a: ĐKXĐ: x<=4/5

\(\sqrt{4-5x}=12\)

=>\(4-5x=12^2=144\)

=>5x=4-144=-140

=>x=-28(nhận)

b: \(\sqrt{4\left(1-x\right)^2}-6=0\)

=>\(2\cdot\left|x-1\right|-6=0\)

=>2|x-1|=6

=>|x-1|=3

=>\(\left[\begin{array}{l}x-1=3\\ x-1=-3\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=4\\ x=-2\end{array}\right.\)

c: \(\sqrt{4x^2+4x+1}=6\)

=>\(\sqrt{\left(2x+1\right)^2}=6\)

=>|2x+1|=6

=>\(\left[\begin{array}{l}2x+1=6\\ 2x+1=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=5\\ 2x=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac52\\ x=-\frac72\end{array}\right.\)

Bài 3:

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC\)

=>\(HC=\frac{3^2}{2}=4,5\left(\operatorname{cm}\right)\)

BC=BH+CH=2+4,5=6,5(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AB^2=BH\cdot BC=2\cdot6,5=13\)

=>\(AB=\sqrt{13}\) (cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(CA^2=CH\cdot CB=4,5\cdot6,5=\frac92\cdot\frac{13}{2}=\frac{117}{4}\)

=>\(CA=\sqrt{\frac{117}{4}}=\frac{3\sqrt{13}}{2}\left(\operatorname{cm}\right)\)

PTHH: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(\left\{{}\begin{matrix}m_{H_2SO_4}=588\cdot5\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,3}{3}\) \(\Rightarrow\) Al₂O₃ còn dư

\(\Rightarrow n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)=n_{Al_2O_3\left(dư\right)}\)

\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1\cdot342}{20,4+588-0,1\cdot102}\cdot100\%\approx5,72\%\)