Câu 5:

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

             a_____3a______a______ \(\dfrac{3}{2}a\)   (mol)

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

             b_____2b_____b______b          (mol)

a) Ta lập được hệ phương trình: \(\left\{{}\begin{matrix}27a+56b=5,5\\\dfrac{3}{2}a+b=\dfrac{4,48}{22,4}=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1\cdot27}{5,5}\cdot100\%\approx49,1\%\\\%m_{Fe}=50,9\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{HCl}=3n_{Al}+2n_{Fe}=0,4\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,4}{0,5}=0,8\left(l\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{FeCl_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{AlCl_3}}=\dfrac{0,1}{0,8}=0,125\left(M\right)\\C_{M_{FeCl_2}}=\dfrac{0,05}{0,8}=0,0625\left(M\right)\end{matrix}\right.\)

Câu 6: 

a+b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\end{matrix}\right.\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

              0,15___0,15_____0,15___0,15  (mol)

            \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

              0,25___0,25____0,25____0,25    (mol)

\(\Rightarrow m_{Fe}=0,25\cdot56=14\left(g\right)\)

c) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) H₂ còn dư, CuO p/ứ hết

\(\Rightarrow n_{Cu}=0,3\left(mol\right)\) \(\Rightarrow m_{Cu}=0,3\cdot64=19,2\left(g\right)\)

1.C

2.D

3.C

4.C

5.C

6.C

7.D

8.D

9.C

10.D

11.A

12.A

5a.

Pt có 2 nghiệm pb lhi:

\(\Delta=9+4m>0\Leftrightarrow m>-\dfrac{9}{4}\)

b. Phương trình có 2 nghiệm khi:

\(\Delta=1+4\left(-2m+1\right)\ge0\Rightarrow m\le\dfrac{5}{8}\)

6.

a. Pt có 2 nghiệm khi: 

\(\Delta'=1-\left(m+2\right)\ge0\Leftrightarrow m\le-1\)

6b

Khi \(m\le-1\), theo hệ thức Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m+2\end{matrix}\right.\)

\(x^2_1+x^2_2=10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\)

\(\Leftrightarrow4-2\left(m+2\right)=10\)

\(\Leftrightarrow m=-5\)

B.

\(x^2_1+x_2^2+4x_1x_2=0\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+2x_1x_2=0\)

\(\Leftrightarrow4+2\left(m+2\right)=0\)

\(\Leftrightarrow m=-4\)

C.

Để biểu thức đề bài xác định \(\Rightarrow x_1x_2\ne0\Rightarrow m\ne-2\)

Khi đó:

\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-3\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=-3\)

\(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=-3x_1x_2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow4+m+2=0\)

\(\Leftrightarrow m=-6\)

D.

\(x_1^3+x_2^3=15\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=15\)

\(\Leftrightarrow8-6\left(m+2\right)=15\)

\(\Leftrightarrow m=-\dfrac{19}{6}\)

Bann ơi tôi xin đề trách nhiệm công nghệ đó với

5. a) Fe + H2SO4 --------> FeSO4 + H2

Mg + H2SO4 --------> MgSO4 + H2

Gọi x, y lần lượt là số mol của Fe, Mg

Ta có \(\left\{{}\begin{matrix}56x+24y=12\\x+y=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,15\\y=0,15\end{matrix}\right.\)

=> \(\%m_{Fe}=\dfrac{0,15.56}{12}.100=70\%\)

=> %m_Mg = 100 -70 = 30%

b) Ta có n _H2SO4 = n _Fe + n_Mg = 0,15 + 0,15 =0,3 (mol)

=>a = \(CM_{H_2SO_4}=\dfrac{0,3}{0,2}=1,5M\)

c) \(CM_{FeSO_4}=\dfrac{0,15}{0,2}=0,75M\)

\(CM_{MgSO_4}=\dfrac{0,15}{0,2}=0,75M\)

6. a) 2Al + 6HCl → 2AlCl3 + 3H2

Zn  + 2HCl →ZnCl2 + H2

Gọi x, y lần lượt là số mol của Al, Zn

\(\left\{{}\begin{matrix}27x+65y=11,9\\\dfrac{3}{2}x+y=\dfrac{8,96}{22,4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

=> \(\%m_{Al}=\dfrac{27.0,2}{11,9}.100=45,38\%\)

=> %m _Zn = 100- 45,38=54,62%

b) \(a=C\%_{HCl}=\dfrac{\left(0,2.2+0,1.2\right).36,5}{200}.100=10,95\%\)

c) \(m_{ddsaupu}=11,9+200-\left(0,2.2+0,1.2\right)=211,3\left(g\right)\)

=> \(C\%_{AlCl_3}=\dfrac{0,2.133,5}{211,3}.100=12,64\%\)

\(C\%_{ZnCl_2}=\dfrac{0,1.136}{211,3}.100=6,44\%\)