Bài 1: 

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,4\cdot36,5}{14,6\%}=100\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)

Bài 2:

PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{KOH}=\dfrac{100\cdot11,2\%}{56}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) H₂SO₄ còn dư, KOH p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_4}=0,1\cdot174=17,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{ddKOH}+m_{ddH_2SO_4}=250\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{17,4}{250}\cdot100\%=6,96\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{250}\cdot100\%=1,96\%\end{matrix}\right.\)

Em đăng sang môn Hoá nha

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)

a)

$Zn + H_2SO_4 \to ZnSO_4 + H_2$
b)

Ta thấy :

$n_{Zn} = \dfrac{2,6}{65} = 0,04 < n_{H_2SO_4} = \dfrac{200.2,45\%}{98}= 0,05$ nên $H_2SO_4$ dư

$n_{H_2} = n_{Zn} = 0,04(mol)$
$V_{H_2} = 0,04.22,4 = 0,896(lít)$

c)

Dung dịch X chứa :

$n_{ZnSO_4} = n_{Zn} = 0,04(mol) \Rightarrow m_{ZnSO_4} = 0,04.161 = 6,44(gam)$
$n_{H_2SO_4\ dư} = 0,05 - 0,04 = 0,01(mol) \Rightarrow m_{H_2SO_4} = 0,01.98 = 0,98(gam)$

d)

Cách 1 : 

$m_{dd} = 2,6 + 200 - 0,04.2 = 202,52(gam)$

Cách 2 : 

Trong dd $H_2SO_4$ : $m_{H_2O} = 200 - 0,05.98 = 195,1(gam)$

Suy ra:
$m_{dd} = m_{chất\ tan} + m_{H_2O} = 6,44 + 0,98 + 195,1 = 202,52(gam)$

e)

$C\%_{H_2SO_4} = \dfrac{0,98}{202,52}.100\% = 0,48\%$
$C\%_{ZnSO_4} = \dfrac{6,44}{202,52}.100\% = 3,18\%$

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

             0,1---->0,3

          Zn + H₂SO₄ --> ZnSO₄ + H₂

           0,3<--------------------0,3

=> m = 0,3.65 = 19,5 (g)

a. \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

   0.25     0.5                       0.25

\(n_{HCl}=\dfrac{18.25}{36.5}=0.5mol\)

\(m_{Zn}=0.25\times65=16.25g\)

\(V_{H_2}=0.25\times22.4=5.6l\)

b. Gọi công thức tổng quát của oxit kim loại R là \(R_2O_n\)

 \(R_2O_n+nH_2\underrightarrow{t^o}2R+nH_2O\)

  \(\dfrac{0.25}{n}\)     0.25

\(M_{oxit}=\dfrac{18}{\dfrac{0.25}{n}}\)

Với n = 1: M = 72 \(\Rightarrow\) Công thức của oxit là FeO: Sắt(II) oxit

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ m_{Cu}=9,7-6,5=3,2g\\ b)C_{\%ZnCl_2}=\dfrac{0,1.136}{6,5+120-0,1.2}\cdot100=10,77\%\)