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

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

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

Bài 2: 

a: \(\Leftrightarrow4x^2-20x+25-4x^2+12x=0\)

=>-8x=-25

hay x=25/8

1. 

a) x (x - 5) + (x + 3)(x - 3)=

= x^2 - 5x + (x + 3)(x - 3)

= x^2 - 5x + x^2 - 9

= x^2 + x^2 - 5x - 9

= 2x^2 - 5x - 9.

b. không thể nhìn thấy hết bài được. Nó bị mất dấu!!

c. (20x^2 + 7x - 6) : (5x - 2)

= (5x - 2) (4x + 3) : (5x - 2)

= 4x + 3.

2. 

a. (2x - 5)^2 - 4x (x - 3)= 0

                       -8x + 25= 0

                -8x + 25 - 25= 0 - 25

                               -8x= -25

                          -8x : 8= -25 : 8

                                 x = 25/8

Vậy x= 25/8

b. 2(x - 5) - x^2 - 5x= 0

                        -10x= 0

              -10x : (-10)= 0 : (-10)

                              x= 0

Vậy x= 0

c. Lí do cũng giống câu b bài 1.

\(d,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}=2+\sqrt{x+1}\\ \Leftrightarrow x-1=2+x+1+4\sqrt{x+1}\\ \Leftrightarrow4\sqrt{x+1}=-4\Leftrightarrow x\in\varnothing\left(4\sqrt{x+1}\ge0\right)\\ g,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}=2\\ \Leftrightarrow2x+2\sqrt{x^2-2x+1}=2\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=\dfrac{2-2x}{2}=1-x\\ \Leftrightarrow\left|x-1\right|=1-x\\ \Leftrightarrow\left[{}\begin{matrix}x-1=1-x\left(x\ge1\right)\\x-1=x-1\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x\in R\end{matrix}\right.\)