a. \(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}=\dfrac{x+100}{94}+\dfrac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

c. \(\Leftrightarrow3x^2+3x-x-1=0\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow\left[\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

a.

\(n_{H_2}=0,3mol;n_{Fe_2O_3}=0,15mol\)

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

0,1-----------0,3-------0,2

\(\dfrac{n_{H_2}}{3}< \dfrac{n_{Fe_2O_3}}{1}\Rightarrow\) \(Fe_2O_3\) dư

\(n_{Fe_2O_3\left(pu\right)}=0,1mol\Rightarrow n_{Fe_2O_3\left(du\right)}=0,15-0,1=0,05mol\)\(\Rightarrow m_{Fe_2O_3\left(du\right)}=0,05.160=8g\)

b. \(m_{Fe}=0,2.56=11,2g\)

c.

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2------------------------------0,2

Hao hụt 20%--> hiệu suất h=80%\(\Rightarrow m_{H_2}=0,2.2.\dfrac{80}{100}=0,32g\)

\(n_{Fe}=0,2mol\)

\(Fe_xO_y+yH_2\rightarrow xFe+yH_2O\)

\(\dfrac{0,2}{x}\)..........................0,2

\(M_{Fe_xO_y}=\dfrac{16}{\dfrac{0,2}{x}}=80x\Leftrightarrow56x+16y=80x\Leftrightarrow\dfrac{x}{y}=\dfrac{2}{3}\)

\(\Rightarrow x=2;y=3\)

Oxit sắt là: \(Fe_2O_3\)

a. Gọi x là số mol Cu--> số mol Fe là 3x

Ta có: \(m_{Fe}+m_{Cu}=23,2\Leftrightarrow56.3x+64.x=23,2\) \(\Rightarrow x=0,1\Rightarrow n_{Cu}=0,1mol;n_{Fe}=0,3mol\)

\(Fe_3O_4+4H_2\rightarrow3Fe+4H_2O\)

0,1---------0,4--------0,3

\(CuO+H_2\rightarrow Cu+H_2O\)

0,1-------0,1------0,1

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2g;m_{CuO}=0,1.80=8g\)

\(n_{H_2}=0,4+0,1=0,5mol\)\(\Rightarrow V_{H_2}=0,5.22,4=11,2l\)

b.

\(3Fe+2O_2\rightarrow Fe_3O_4\)

0,3-------0,2

\(Cu+O_2\rightarrow CuO\)

0,1-----0,1

\(\Rightarrow n_{O_2}=0,2+0,1=0,3mol\Rightarrow V_{O_2}=0,3.22,4=6,72l\)

\(\Rightarrow V_{kk}=5.V_{O_2}=33,6l\)

c.

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(\dfrac{1}{3}\)---------------------------------------0,5

Hao hụt 30%--> hiệu suất h = 70% \(\Rightarrow m_{Al}=\dfrac{1}{3}.27.\dfrac{100}{70}\simeq12,86g\)

\(\dfrac{43}{30}=\dfrac{30+13}{30}=\dfrac{30}{30}+\dfrac{13}{30}=1+\dfrac{1}{\dfrac{30}{13}}\)

\(=1+\dfrac{1}{\dfrac{26+4}{13}}=1+\dfrac{1}{\dfrac{26}{13}+\dfrac{4}{13}}=1+\dfrac{1}{2+\dfrac{4}{13}}\)

\(=1+\dfrac{1}{2+\dfrac{1}{\dfrac{13}{4}}}=1+\dfrac{1}{2+\dfrac{1}{\dfrac{12+1}{4}}}=1+\dfrac{1}{2+\dfrac{1}{\dfrac{12}{4}+\dfrac{1}{4}}}=1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}\)

Vậy \(x=2;y-3;z=4\)

A lớn nhất\(\Leftrightarrow\left[\left(x+2\right)^2+5\right]\) nhỏ nhất

Ta có: \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+5\ge5\)

\(\Rightarrow\)\(\left[\left(x+2\right)^2+5\right]\)nhỏ nhất = 5 khi \(x+2=0\Leftrightarrow x=-2\)

Vậy: \(A\)lớn nhất =\(\dfrac{10}{5}=2\)

\(\Leftrightarrow2015^{2x^2-4x}=2015^0\)\(\Leftrightarrow2x^2-4x=0\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow x=0\) hay \(x-2=0\)\(\Leftrightarrow x=0\) hay x = 2

a, b là số mol X và Y

\(n_{O_2}=0,55mol;n_{CO_2}=0,35\)

\(C_nH_{2n+2}+\left(\dfrac{3n+1}{2}\right)O_2\rightarrow nCO_2+\left(n+1\right)H_2O\)

a--------------\(\dfrac{3n+1}{2}a\) ------------\(na\)

\(C_mH_{2m-2}+\left(\dfrac{3m-1}{2}\right)O_2\rightarrow mCO_2+\left(m-1\right)H_2O\)

b---------------\(\dfrac{3m-1}{2}b\)------------\(mb\)

Ta có:

\(\left\{\begin{matrix}a+b=0,25\\\dfrac{3n+1}{2}a+\dfrac{3m-1}{2}b=0,55\\na+mb=0,35\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}a+b=0,25\\a-b=0,05\\na+mb=0,35\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}a=0,15\\b=0,1\\3n+2m=7\end{matrix}\right.\)

\(\Rightarrow n=1;m=2\)\(\Rightarrow X:CH_4;Y:C_2H_2\)

\(n_{H_2SO_4}=0,9.0,2=0,18mol\)

\(H_2SO_4\) dư 20%\(\Rightarrow n_{H_2SO_4\left(pu\right)}=0,18.\dfrac{100}{120}=0,15mol\) ; \(n_{H_2}=\dfrac{0,672}{22,4}=0,03mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,02----0,03----------------------------0,03
\(Fe_xO_y+yH_2SO_4\rightarrow Fe_x\left(SO_4\right)_y+yH_2O\)

\(\dfrac{0,12}{y}\) ---------- 0,12

Ta có: \(m_{hh}=6,94\Leftrightarrow\left(56x+16y\right).\dfrac{0,12}{y}+0,02.27\)

\(\Leftrightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow Fe_2O_3\)

\(\%Al=\dfrac{0,2.27}{6,94}.100\%\simeq77,8\%\Rightarrow\%Fe_2O_3=22,2\%\)