1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

\(=\dfrac{2x^2y^2}{3xy^2}-\dfrac{2ax+3x}{3a}=\dfrac{2x}{3}-\dfrac{2ax+3x}{3a}\)

\(=\dfrac{2xa-2xa-3x}{3a}=\dfrac{-3x}{3a}=-\dfrac{x}{a}\)

\(=\dfrac{5}{3}-\dfrac{5a-6}{3a}=\dfrac{5a-5a+6}{3a}=\dfrac{6}{3a}=\dfrac{2}{a}\)

\(=\dfrac{2x-3a}{2a}+\dfrac{3}{2}=\dfrac{2x-3a+3a}{2a}=\dfrac{2x}{2a}=\dfrac{x}{a}\)

\(=\dfrac{2-a}{2a}+\dfrac{1}{2x}=\dfrac{4x-2xa+2a}{4xa}=\dfrac{2x-xa+a}{xa}\)

a) \(n_{Fe}=\dfrac{28}{56}=0.5\left(mol\right)\)

\(n_{O_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

     \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)

\(Bđ:0.5....0.2\)

\(Pư:0.3.....0.2........0.1\)

\(Kt:0.2.......0..........0.1\)

\(m_{Fe\left(dư\right)}=0.2\cdot56=11.2\left(g\right)\)

\(m_{Fe_3O_4}=0.1\cdot232=23.2\left(g\right)\)

a. \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

\(n_{O_2}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)

Ta thấy : 0,5 > 0,2 => Fe dư , O₂ đủ

PTHH : 3Fe + 2O₂ ---t^o---> Fe₃O₄

             0,3      0,2                0,1

\(m_{Fe\left(dư\right)}=\left(0,5-0,3\right).56=11,2\left(g\right)\)

b. \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

bn đg thi hả